18.757 (representation of lie algebras) lecture notes · 18.757 (representation of lie algebras)...

18.757 (Representation of Lie Algebras) Lecture NotesMassachusetts Institute of Technology

Evan Chen

Spring 2016

This is MIT’s graduate course 18.757, instructed by Laura Rider. Theformal name for this class is “Representations of Lie Algebras”.

The permanent URL for this document is http://web.evanchen.cc/

coursework.html, along with all my other course notes.

Contents

1 February 2, 2016 4§1.1 Representations of groups . . . . . . . . . . . . . . . . . . . . . . . . . . 4§1.2 Digression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4§1.3 More on representations . . . . . . . . . . . . . . . . . . . . . . . . . . . 5§1.4 Recovering G . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5§1.5 Schur lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5§1.6 Lie groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 February 4, 2016 7§2.1 Representations for Lie groups . . . . . . . . . . . . . . . . . . . . . . . . 7§2.2 Measure theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

3 February 9, 2016 10§3.1 Matrix coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10§3.2 Bimodule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11§3.3 Characters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

4 February 11, 2016 14§4.1 Left/right action . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14§4.2 Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14§4.3 Complexification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14§4.4 Representations of sl2 C . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

5 February 18, 2016 17§5.1 An algebraic lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17§5.2 Universal enveloping algebra . . . . . . . . . . . . . . . . . . . . . . . . . 18§5.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

6 February 23, 2016 20§6.1 Hopf algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20§6.2 Monoidal functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

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http://web.evanchen.cc/coursework.html

http://web.evanchen.cc/coursework.html

Evan Chen (Spring 2016) Contents

§6.3 Application to k[G] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22§6.4 Centers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

7 February 25, 2016 25§7.1 Brief remark on monoidal categories . . . . . . . . . . . . . . . . . . . . 25§7.2 Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25§7.3 Solvable and nilpotent algebras . . . . . . . . . . . . . . . . . . . . . . . 25§7.4 Representation theory of solvable Lie algebras . . . . . . . . . . . . . . . 28

8 March 1, 2016 29§8.1 Lie’s theorem and Engel’s theorem . . . . . . . . . . . . . . . . . . . . . 29§8.2 Jordan decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30§8.3 Jordan decomposition of Lie elements . . . . . . . . . . . . . . . . . . . . 31§8.4 Toral subalegbras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

9 March 3, 2016 34§9.1 Review of varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34§9.2 Algebraic groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35§9.3 Group actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

10 March 8, 2016 38§10.1 Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38§10.2 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38§10.3 Quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39§10.4 Complete varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40§10.5 Parabolic subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

11 March 10, 2016 42§11.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42§11.2 Reductive Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42§11.3 Killing form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43§11.4 Compact Lie groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44§11.5 One-dimensional representations . . . . . . . . . . . . . . . . . . . . . . . 44§11.6 Borel and parabolic subgroups . . . . . . . . . . . . . . . . . . . . . . . . 45

12 March 15, 2016 46§12.1 Proof of proposition from last time . . . . . . . . . . . . . . . . . . . . . 46§12.2 Rational representations . . . . . . . . . . . . . . . . . . . . . . . . . . . 47§12.3 Diagonalizable groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

13 March 17, 2016 50§13.1 More on diagonalizable groups . . . . . . . . . . . . . . . . . . . . . . . . 50§13.2 Rigidity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51§13.3 Tori . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52§13.4 Parabolic subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52§13.5 Borel subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

14 March 29, 2016 53§14.1 Exercises for Tuesday April 5 . . . . . . . . . . . . . . . . . . . . . . . . 53§14.2 Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53§14.3 Lie-Kolchin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

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Evan Chen (Spring 2016) Contents

15 March 31, 2016 57§15.1 Maximal Torus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57§15.2 Main idea . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57§15.3 Proof of main theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59§15.4 An example! . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59§15.5 Cartan subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59§15.6 Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

16 April 5, 2016 60§16.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60§16.2 More on maximal tori . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60§16.3 Maximal tori are conjugate . . . . . . . . . . . . . . . . . . . . . . . . . . 61

17 April 14, 2016 64§17.1 Vector bundles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64§17.2 Interlude . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64§17.3 Weyl group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

18 April 21, 2016 66§18.1 Setup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66§18.2 Euclidean reflections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

19 May 3, 2016 68§19.1 Some functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68§19.2 Root systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

20 May 12, 2016 71

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Evan Chen (Spring 2016) 1 February 2, 2016

§1 February 2, 2016

Following Kirillov, chapters 4 and 8. Office is 2-246a, office hours are Tu 11-12 and W11:30-12:30.

Exercises: 2.1-2.4, 2.12, 2.13[K]; choose 3 to submit.

§§1.1 Representations of groups

Definition 1.1. Let G be a group. A complex representation of G is a complexvector space V with group homomorphism ρ : G→ Aut(V ) = GL(V ).

Example 1.2 (Examples of Representations)

(a) For any G, V = C, we have a trivial representation G 7→ 1 ∈ C×.

(b) Let G = S2 = {e, τ}. Then the sign representation S2 → C× is e 7→ 1 andτ 7→ −1.

(c) Let Sn be the symmetric group on n letters. Let V = Cn with basis b1, . . . , bn.Then Sn → GL(V ) by permuting the basis and extending linearly.

Definition 1.3. A morphism of representations f : V →W is a linear map commutingwith the action of G: fρV (g) = ρW (g)f . We denote by HomG(V,W ) the set of suchmorphisms. We let RepG denote finite dimensional complex representations.

Definition 1.4. A subrepresentation W of V is a linear subspace such that ρ(g)W ⊆W for all g.

Definition 1.5. Define the quotient representation V/W in the obvious way.

Thus, short exact sequence of representations

0→W → V → V/W → 0.

If f : V → W , im f is a subrepresentation, so is V/ ker f , and these are isomorphic asG-representations.

Remark 1.6. What we’re getting at is that RepG is an abelian category. To do thiswe’d need to define V ⊕W (obvious way)

§§1.2 Digression

A student wants to make cohomology happen. In order to do this, we need a functor(derived functors). One example would be

F : Rep(G)→ VectC

but this isn’t a good example since it’s exact. More functors:

• HomG(V,−)

• EndG(V )

• HomG(C,−)

• (−)G (G-invariants).

4

https://en.wikipedia.org/wiki/Derived_functor


§§1.3 More on representations

Definition 1.7. A representation V is irreducible if its only subrepresentations are 0and V .

In categorical language this is always called “simple”.

Example 1.8

The group S3 acting on GL(C3) is not irreducible because the diagonal (t, t, t) is asubrep. The set of (a, b, c) with a+ b+ c = 0 is also a subrep; it is two-dimensionaland one can check its irreducible.

§§1.4 Recovering G

Can we reconstruct G from its category of representations? If so, the structure of G hasmultiplication, id and inversion, so we want to have a representation analogues of theseoperations.

Definition 1.9. If V,W ∈ Rep(G) then define V ⊗W in the obvious way: v ⊗ w 7→ρV (g)v ⊗ ρW (g)w.

Observe that the trivial representation C is now the identity; we have C⊗ V ∼= V .Inversion is less straightforward; if dimV ≥ 2 then dimV ⊗W 6= 1. On the other

hand,

Definition 1.10. If V ∈ Rep(G) we define the dual V ∨ in the obvious way.

Then we at least have a map V ⊗ V ∨ ∈ C.

Remark 1.11. If G is finite, then we can get C[G] by⊕

V ∈Rep(G) V ⊗ V ∨ But one can’tactually read off G from C[G] easily.

Remark 1.12. Mark Sellke points out that D8 and quaternions Q8 have the samecharacter table, so the answer to the question is actually negative.

There is a natural faithful forgetful functor RepG→ VectC. Adjoint?Suppose we have a fully general C , abelian category with a symmetric ⊗. We’d like to

factor it through a diagram

C ,⊗ ?- RepG

Vect(C)

Forget

?�

Here the forgetful functor is a fiber functor (exact, faithful to Vect(C)).

§§1.5 Schur lemma

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Lemma 1.13 (Schur’s lemma)

Let V and W be irreducible representations of G. Then

HomG(V,W ) =

{C · id V ∼= W

0 else.

In particular, over C any morphism V → V is constant, when V is irreducible since ithas an eigenspace.

Goal for course: understand RepG, its irreducible representations, and when under-standing the irreducibles is enough to get all of RepG. (Answer to “enough”: semisimpleor completely reducible.)

Next class, we answer this question for finite groups, and see for Lie groups G anexample of an indecomposable but reducible rep.

§§1.6 Lie groups

Let G be a group. It’s a real Lie group if G is also a smooth manifold such thatmultiplication G×G→ G and inversion G→ G are smooth.

A representation of a Lie group G is one such that ρ : G→ GL(V ) is also a morphismof Lie groups.

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First homework due February 11, 2016Added: 4.3-4.5, 4.11, 4.13, choose 2 from Tuesday and 2 from these

§§2.1 Representations for Lie groups

Example 2.1 (Examples of Lie groups)

(a) R,+

(b) R×,×, disconnected, with identity component R>0,×

(c) S1

(d) SU(2), matrices A with A ∈ GL2 C, detA = 1.

Exercise 2.2. Check that SU(2) can be identified with S3. Thus compact, simplyconnected.

The complex Lie group GLnC acts on Cn. This representation is irreducible (why?).If f : Cn → Cn is a G-map then f = λid by Schur. So the center of GLnC is the scalarsC×.

Similarly the center of SLnC is the nth roots of unity.Similarly, the unitary group U(n) has S1 as its center.

Proposition 2.3

Now let G be an abelian group. Then all irreps of G are one-dimensional.

Proof. Napkin problem 45B.

Here is an indecomposable but not irreducible representation over characteristic zero.Let ρ : (R,+)→ GL2 C by

ρ(t) =

(1 t0 1

).

The span of (1, 0) is a subrepresentation. So there is an exact sequence

0→ C ↪→ C⊕2 � C→ 0.

But it turns out this doesn’t split.Goal:

• How to decompose representations?

• What are sufficient conditions on G such that we can always split into direct sumof irreducibles?

Lemma 2.4

Let (V, ρ) be a representation of G. If A : V → V is intertwining, then Vλ is asubrepresentation for every λ.

In particular, if A is diagonalizable, then V =⊕Vλ as G-reps.

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Corollary 2.5

If g ∈ Z(G), and ρ(g) is diagonalizable then V =⊕Vλ.

Definition 2.6. A complex rep V is unitary if there exists an inner product which isG-invariant.

Theorem 2.7

Finite-dimensional unitary representations are completely reducible.

Proof. If W ⊆ V is a subrepresenattion, then so is the orthogonal complement W⊥.

Theorem 2.8

A representation of a finite group is unitary.

Proof. If B is a linear form then

B(v, w) =∑g∈G

B(gv, gw)

is a G-invariant positive definite inner form.

In order to get this to work with a Lie group, we want to replace∑

g∈G with∫g.

§§2.2 Measure theory

Let X be a topological space, locally compact and Hausdorff. Then a σ-algebra Σ ⊆ P(X)is a family of sets which are closed under complements, countable unions, and intersections.

The measure µ should satisfy the obvious additive axioms and also µ(C) <∞ for anycompact C. It should also be inner and outer regular, meaning

µ(U) = sup {µ(K) | K ⊆ U compact}

and similarly for U ⊆ K.

Definition 2.9. If G is a topological group a left Haar measure on G is a Borelmeasure µ on G such that µ(gA) = A i.e. it is translation invariant. (A Borel measure isa measure which also contains every open set.)

For example, the Lebesgue measure on R which sends µ([a, b]) = b− a is translationinvariant on R, but of course not on R×.

Anyways, we want to define a Borel measure on G. The idea is that if G is compactand we have a “small” compact subset K of G, then we can use the G-action to translateK around and then use it to define the measure on all of G. This is what we’ll do.

Let K be compact, and let V ⊆ G be a set with nonempty interior V (for examplean open set). Then K is covered by {gV | g ∈ G}. Now K is compact, so it has a finitesubcover, and we define (K : V ) to be the size of the smallest such cover.

Now, for any U we define

µu(K) =(K : U)

(K0 : U)

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where K0 is a particular fixed compact subset with nonempty interior. It’s easy to see0 ≤ µU (K) ≤ (K : K0).

Finally, define

X =∏

G⊇K compact

[0, (K : K0)].

The space X is compact under the product topology by Tychonoff’s theorem. Also, forany U we can regard µU as a point in X, namely the point (µU (K))K⊆G.

Now if V is open we can define the subset of the space

C(V ) = {µU | V ⊇ U open} ⊆ X

a closed subset of X. This has the finite intersection property for Vi ⊆ G, because

µ⋂Vi ∈⋂i

C(Vi)

across i = 1, . . . , n. Since X is compact, this now implies that⋂G⊇V open

C(V ) 6= ∅

and we take any µ in this set. This µ is a Radon measure on G.One can check that this µ satisfies all the properties we want, but we at least get

translation invariance for free, since (K : U) = (gK : U).What we have just constructed is:

Theorem 2.10 (Haar, Weil, Cartan)

If G is a (Hausdorff) locally compact topological group then there exists a left Haarmeasure, unique up to some positive scalar.

Remark 2.11. There is a precise way to get a right Haar measure from a left Haarmeasure and vice-versa. If G is compact, we can declare µ(G) = 1 which will uniquelyidentify the measure we want.

Remark 2.12. In a compact group, in fact the left and right invariant Haar measurescoincide.

Armed with the Haar measure, we have

Theorem 2.13

Any finite dimensional representation of a compact Lie group is unitary and henceis completely reducible.

Proof. Same as in the group case: if B : V × V → C is any inner product we define theaveraged inner form by

B(v, w) =

∫GB(gv, gw) dg

where dg is the right-invariant Haar measure. Then the G-invariance of B follows.

Remark 2.14. Assume the special case that G is a compact Lie group. It is a manifold,and one can show its orientable. Thus there is volume form on G, and then we can showthat this volume form is G-invariant.

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Last time: representations of compact Lie groups are unitary, and hence semisimple.

Example 3.1

Let S1 = R/Z be the circle group. We compute its irreducible representations. LetVk be the irreducible representation of S1 on C by z 7→ zk ∈ C×.

For the rest of today, assume G is a compact Lie group.

§§3.1 Matrix coefficients

Assume G compact, and write ρ(g) ∈ GL(V ) as a matrix in some basis of V . Thus wehave maps G→ C for each i, j.

It’s more elegant to do this with an inner form.

Definition 3.2. Let φ : G→ GL(V ) be a group representation, and define the matrixcoefficient by

φu,v : G→ C by g 7→ 〈φ(g)u, v〉

for every u, v ∈ V . (Here V has a specific G-invariant 〈−,−〉.)

Classically, we wanted to study C∞(G,C), so we endow it with an inner product∫Gf(x)g(x) dx.

Our goal is to get inner product relations along matrix coefficients.

Lemma 3.3

Let ` : V ′ → V be a map of representations (V ′, φ′) and (V, φ). Define the operator

L =

∫Gφ(x)`φ′(x−1) dx.

Then L : V ′ → V is intertwining In particular,

• L = 0 if V 6∼= V ′ is irreducible.

• L = λid if V ∼= V ′ is irreducible.

Proof. Schur’s lemma.

Theorem 3.4

Suppose V and V ′ are non-isomorphic irreps of G.

(a) The matrix coefficients are pairwise orthogonal in C∞(G).

(b) Let φ : G→ GLV be irreducible, then

〈φu1v1 , φu2v2〉 =〈u1, v1〉〈u2, v2〉

dimV.

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Proof. (a) Consider u, v ∈ V and u′, v′ ∈ V ′ arbitrary. We will show that φuv and φu′v′

are orthogonal.

If ` : V ′ → V by w′ 7→ 〈w′, u′〉u, then we know that L = 0, meaning

0 =⟨Lv′, v

⟩=

∫G

⟨φ(x)`(φ′(x−1)v′), v

⟩dx

=

∫G

⟨φ(x)

⟨φ′(x−1)v′, u′

⟩u, v⟩dx

=

∫G

⟨φ′(x−1)v′, u′

⟩〈φ(x)u, v〉 dx

=

∫G〈φ(x)u, v〉〈u′, φ′(x−1)v′〉 dx

=

∫G〈φ(x)u, v〉〈φ′(x)u′, v′〉 dx

the last step since the inner form is G-invariant.

(b) Again let ` : V → V by w 7→ 〈w, u2〉u1. Note that u1 is the only nonzero eigenvectorof `, and its eigenvalue is 〈u1, u2〉. So Tr ` = 〈u1, u2〉.As L : V → V is a constant map L = λid, and we have TrL = Tr ` = 〈u1, u2〉, itfollows that L is multiplication by

λ =〈u1, u2〉dimV

.

By the same argument in part (a) we have∫G〈φ(x)u1, v1〉〈φ(x)u2, v2〉 dx = 〈Lv2, v1〉 .

By the work above, we deduce

〈Lv2, v1〉 =〈u1, u2〉dimV

〈v2, v1〉 =〈u1, u2〉dimV

〈v1, v2〉.

§§3.2 Bimodule

We defined a map V × V → C∞(G) by (v, w) 7→ (g 7→ (φ(g)v, w). We can generalize thisto a map

m : V ∨ ⊗ V → C∞(G) by ξ ⊗ v 7→ ξ (ρ(g)v) .

Equivalently, define an “inner product” on V ∨ ⊗ V by 〈ξ, v〉 = ξ(v). Hence above map isξ ⊗ v 7→ 〈ξ, ρ(g)v〉.

Then, we equip V ∨ ⊗ V with the inner product

〈ξ ⊗ v, η ⊗ w〉 =1

dimV〈ξ, η〉〈v, w〉 .

We can now endow it with the following G-bimodule structure:

• Left action: ξ ⊗ (ρ(g)v).

• Right action: (ξ ◦ ρ(g))⊗ v.

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Meanwhile, we endow C∞(G,C) with the following action:

• Left action: (g · f)(h) = f(g−1h).

• Right action: (f · g)(h) = f(gh).

Definition 3.5. Let G be the set of irreps of G up to isomorphism.

Theorem 3.6

(1) m is a morphism of G-bimodules.

(2) m preserves inner products.

(3) m is injective.

Proof. (1) is computation, (2) follows from earlier theorem with some adjustment, and(3) follows by (2).

Thus, m defines a mapp

m :⊕V ∈G

V ∨ ⊗ V → L2(G, dx).

The image of this map is dense, and in fact if we take the completed direct sum this mapis an isomorphism (Peter-Weyl).

§§3.3 Characters

Let (V, φ) be a rep of G. Then the character of φ is

χφ(x) = Trφ(x).

Properties:

1. The trivial representation has character identically 1.

2. χV⊕W = χV + χW .

3. χV⊗W = χV χW .

4. χV (ghg−1) = χV (h).

5. χV ∨ = χV .

We can also use the inner product in C∞(G).

Theorem 3.7 (Orthogonality of characters)

Let V and W be irreps of G. Then

〈χV , χW 〉 =

{1 V ∼= W

0 V 6∼= W.

Thus {χV | V ∈ G} is an orthonormal basis in the Hilbert space L2(G, dx)G.

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Corollary 3.8

If V is a finite-dimensional representation of G, define

ni = 〈χV , χVi〉 Vi ∈ G

thenV ∼=

⊕i

V ⊕nii .

13



Exercise:

• Chapter 3: 3.3, 3.6, 3.8, 3.13, 3.18

• Chapter 4: 4.11, 4.2, 4.12

§§4.1 Left/right action

The issue last time is that RgRh = Rgh.

• Left action: Lg(ξ ⊗ v) = gξ ⊗ v.

• Right action: Rg(ξ ⊗ v) = ξ ⊗ gv.

• Left action: Lgf(x) = f(g−1x).

• Right action: Rgf(x) = f(xg).

§§4.2 Lie algebras

A Lie algebra g is a K-vector space (K ∈ R,C) equipped with a Lie bracket [, ] : g×g→g.

Remark 4.1. We can make every associate algebra A with unit and construct thebracket [, ] : A×A→ A on it.

A representation of a Lie algebra g is a vector space V and a map of Lie algebrasg→ End(V ). (Thus V is a g-module.)

§§4.3 Complexification

Given real g we can build a complexification gCdef= g ⊗R C. Building this same

complexification on groups is harder.The beauty is that

Theorem 4.2

For G a simply connected Lie group, the category of finite dimensional representationsof G coincides with the category of finite dimensional g-modules.

Example 4.3

(a) If G = S1, then g = R with bracket zero.

(b) Classical groups have Lie algebras. For example, GLn(C) gives gln(C).

(c) sln(C) is the traceless n× n matrices.

LetU(n) =

{M ∈ GLnC |MM † = id

}

14


be the unitary group. Then

u(n) ={M ∈ glnC |M +M † = 0

}.

Now, using the fact that any matrix can be written as a sum of a Skew-Hermitian andHermitian matrix, we get

u(n)C = u(n)⊕ i u(n) ' glnC.

Similarly, if SU(n) the special unitary group, then su(n) is a real Lie algebra, and itscomplexification is sln(C).

Lemma 4.4

Let g be a real Lie algebra and gC its complexification. Then the categories ofg-modules and gC-modules coincide.

Since SU(2) and SL2 C are simply connected, they have the same category of repre-sentations as their Lie algebras by the theorem. Since we saw above that su(2) hascomplexification sl2 C, it follows that

The categories of representations of each of SU(2), su(2), sl2 C, SL2(C) areall the same.

In particular, since SU(2) is compact each of these four groups/algebras is completelyreducible.

§§4.4 Representations of sl2 C

Holy cow I am not doing this a second time. (Etingof’s problem 1.55.) Let sl2(C) havebasis E, F , H as usual.

Theorem 4.5 (All Irreps of sl2 C)

For N ≥ 0, there is a unique (N + 1) dimensional irrep written as

E =

0 N 0 0 . . . 0 00 0 N − 1 0 . . . 0 00 0 0 N − 2 . . . 0 0...

......

.... . .

......

0 0 0 0 . . . 2 00 0 0 0 . . . 0 10 0 0 0 . . . 0 0

F =

0 0 0 . . . 0 0 01 0 0 . . . 0 0 00 2 0 . . . 0 0 00 0 3 . . . 0 0 0...

......

. . ....

......

0 0 0 . . . N − 1 0 00 0 0 . . . 0 N 0

H =

N 0 0 . . . 0 0 00 N − 2 0 . . . 0 0 00 0 N − 4 . . . 0 0 0...

......

. . ....

......

0 0 0 . . . −N + 4 0 00 0 0 . . . 0 −N + 2 00 0 0 . . . 0 0 −N

15


So these Vn are pairwise isomorphic and the only irreducible representations. Moreover,by comments above sl2 C is semisimple, hence all finite-dimensional representations ofsl2 C are sums of these guys.

The notation used by Kirillov is: let V [λ] be the λ-eigenspace of V . In particular, asvector spaces any representation V decomposes as

⊕n V [n] (i.e. one can pick a basis of

integer-valued eigenvectors).We now extend this by formally defining a vector space Mλ with infinite basis v0, v1,

. . . . We view it as an sl2-module with action

hvk = (λ− 2k)vk

fvk = (k + 1)vk+1

evk = (λ− k + 1)vk−1 k > 0

ev0 = 0

Lemma 4.6

Suppose V is a finite-dimensional irreducible sl2-module with nonzero highest weightvector of weight λ. Then there is a surjective map

Mλ � V.

Thus V 'Mλ/W .

Question 4.7. What can you say about the weights of the kernel?

16



Let g be a Lie algebra over K, Our goal is to lift g-representations to representations ofa certain associative algebra.

Definition 5.1. The universal enveloping algebra of g, denoted Ug, is generated bythe tensor algebra

Tg =⊕n≥0

g⊗n

(whose algebra multiplication is ⊗) modulo the relations (a two-sided ideal)

x⊗ y − y ⊗ x− [x, y].

As we all know taking modulo relations makes it hard to describe Ug concretely, butwe want Ug to be interesting anyways. In particular, we would like the canonical map

g ↪→ Ug

to be injective.

§§5.1 An algebraic lemma

Fix an ordered basis {Xi}i∈A. For a sorted sequence I = (i1, . . . , ip) of elements of A,write i ≤ I if i ≤ min{ij}.

Consider the symmetric algebra

Sg = K[zi]

where the zi run through i ∈ A and commute. We can consider it as filtered by degree inthe obvious way, giving

S0 ⊆ S1 ⊆ S2 ⊆ . . .

where Sp is the span of monomials with ≤ p terms in it. (This is a little weaker thangrading.)

Proposition 5.2 (Hard Part of Poincare–Birkoff–Witt)

There is a representation of g on Sg, say π : g→ gl(Sg), such that

(1) π(xi)zI = zizI for i ≤ I.

(2) π(xi)zI − zizI ∈ Sp for all I with zI ∈ Sp.

(3) π(xi) (π(xj)zJ) = (π(xj)π(xi)zJ) + π[xi, xj ]zJ .

Proof. By induction: we define π(xi) on Sp for p inductively, giving π on S.If p = 0, then π(xi)1 = zi.Inductively, for i ≤ I we set

π(xi)zI = zizI

and otherwise if j < i with I = (j, J) then

π(xi)zI = π(xi)(zizJ) = π(xi)π(xj)zJ = π(xj)π(xi)zJ + π[xi, xj ]zJ .

Now, take about four blackboards worth of heavy calculation to verify this works.

17


§§5.2 Universal enveloping algebra

Theorem 5.3 (Universality of Ug)

Ug satisfies the following universal property: for any map ρ : g→ A which satisfiesρ(x)ρ(y) − ρ(y)ρ(x) = ρ([x, y]) there exists a unique algebra homomorphism ρ :Ug→ A which makes the diagram

g Ug

A

ρ

ρ

commute.

Proof. Immediate.

Theorem 5.4 (Poincare–Birkoff–Witt)

fix {Xi}i∈A an ordered basis of g. Then the set of all monomials

i(Xj1i1

)i(Xj2i2

) . . . i(Xjnin

) i1 < · · · < in, and jk ≥ 0

is a basis of Ug.

To do this we first apply the following lemma.

Lemma 5.5 (Re-ordering indices)

If z1, . . . , zp ∈ g and σ is a permutation of {1, . . . , p} then z1 . . . zp − zσ(1) . . . zσ(p) ∈Up−1g.

Proof. Easy. Any permutation is a product of simple transpositions.

Proof of PBW Theorem. It’s easy to see that the image is spanning, since All monomialsspan Tg, so they also span Ug (the previous lemma shows that “increasing” monomialsare sufficient).

Now, we have to show it’s a basis. We take our action g→ gl(Sg) that we worked veryhard to prove. Thus, we can lift it to a representation π : Ug→ GL(Sg), i.e. we now seeSg is a Ug-module. Now, by condition

π(xi1 . . . xip) · 1Sg = zi1 . . . zip ∈ Sg

and these are linearly independent in Sg, so they must be linearly independent in Ug aswell.

§§5.3 Examples

18


Example 5.6

Let G = (R,+). This gives the Lie algebra g = (R, [, ] = 0). We complexify this toget gC = (C, [, ] = 0) and then we can put

U(C, [, ] = 0) = S(C) ∼= C[x].

Now, consider Ctriv. We compute Ext1(Ctriv,Ctriv). Take the following projectiveresolution, call it P •, given by the diagram

0

C[x]

C[x] Ctriv∼= C[x]/(x)

0

·x

and dualize it by Hom(−,Ctriv) to get

0 Hom(C[x],Ctriv) ' C Hom(C[x],Ctriv) ' C 00 .

Thus, Ext1(Ctriv,Ctriv) = C.

Exercise 5.7. Compute Ext1(Ca,Cb) where Cx=a ↔ C[x]/(x− a).

Exercise 5.8. We have HomC[x](P•, P •) is a dg-algebra. What can you say about it?

Can you write down any dg-modules?

Exercise 5.9. We have the cohomology H•(Hom(P •, P •)) ∼= Ext•(Ctriv,Ctriv). Writedown some simple modules.

19



Side remark: it should be possible to distinguish the categories of representations of D8

and Q8 using ⊗-structure.In this lecture, the group G is finite.

§§6.1 Hopf algebras

Recall that an associative k-algebra is a k-vector space A equipped with a map m :A⊗A→ A and i : k ↪→ A (unit).

Then a k-coalgebra is a map

∆ : A→ A⊗A ε : A→ k

called comultiplication and counit respectively. See https://en.wikipedia.org/wiki/

Coalgebra.Now a Hopf algebra A is a bialgebra A over k equipped with a so-called antipode

S : A→ A. We require that the diagram

A⊗A A⊗A

A k A

A⊗A A⊗A

S⊗idA

m∆

∆

ε i

idA⊗S

m

commutes.A group-like element in A is an element of

G = {x ∈ A | ∆(x) = x⊗ x} .

Exercise 6.1. Show that G is a group with multiplication m and inversion S.

Example 6.2 (Group algebra is Hopf algebra)

The group algebra k[G] is a Hopf algebra with

• m, i as expected.

• ε the counit is the trivial representation.

• ∆ comes form g 7→ g ⊗ g extended linearly.

• S takes g 7→ g−1 extended linearly.

Theorem 6.3

The group-like elements are precisely the basis elements 1k · g ∈ k[g].

20

https://en.wikipedia.org/wiki/Coalgebra

https://en.wikipedia.org/wiki/Coalgebra


Proof. Assume V =∑

g∈G agg is grouplike. Then by assumption we should have∑g∈G

ag(g ⊗ g) = ∆(v) =∑g∈G

∑h∈G

agah(g ⊗ h).

Comparing each coefficient, we get that

agah =

{ag g = h

0 otherwise.

This can only occur if some ag is 1 and the remaining coefficients are all zero.

§§6.2 Monoidal functors

A monoidal category or tensor category is a category C equipped with a functor⊗ : C × C → C which is associative up to a natural isomorphism and has an elementwhich is a left/right identity, also up to natural isomorphism. Thus for any A,B,C ∈ Cwe have a natural isomorphism

A⊗ (B ⊗ C)aA,B,C−−−−→ (A⊗B)⊗ C.

Now suppose we have two categories (C ,⊗C ) and (D ,⊗D). Then a monoidal functorF : C → D is a functor for which we additionally need to select an isomorphism

F (A⊗B)tA,B−−−→ F (A)⊗ F (B).

We then require that the diagram

F (A⊗ (B ⊗ C)) F (A)⊗ F (B ⊗ C)

F ((A⊗B)⊗ C) F (A)⊗ (FB ⊗ FC)

F (A⊗B)⊗ FC (FA⊗ FB)⊗ FC

F (aA,B,C)

tA,B⊗C

id⊗tB,C

tA⊗B,C

tA,B⊗id

aFA,FB,FC

commutes, plus some additional compatibility conditions with the identity of the ⊗’s.We also have a notion of a natural transformation of two functors t : F → G, i.e.

making the squares

FA FB

GA GB

tA

Ff

tB

Gf

commute. Now, suppose F : C → C is a monoidal functor. Then an automorphism is anatural isomorphism t : F → F .

Remark 6.4. Categorical remark: when using abstraction we use equivalence of cate-gories which lets us see similarities between different areas. This is weaker than the veryuseless notion of an isomorphism of categories.

21


§§6.3 Application to k[G]

Consider the category of k[G] modules endowed with the monoidal ⊗ (which is just the⊗k representation). We want to reconstruct G from its representations.

Let U be the forgetful functor

U : Modk[G] → Vectk.

It’s easy to see this is in fact an monoidal functor.Now we claim the following:

Theorem 6.5 (G is isomorphic to Aut⊗(U))

Consider the mapi : G→ Aut⊗(U) by g 7→ T g.

Here, the natural transformation T g is defined by the components

T g(V,φ) : (V, φ)→ U(V, φ) = V by v 7→ φ(g)v.

Then i is an isomorphism of groups.

In particular, using only ⊗ structure this exhibits an isomorphism G ∼= Aut⊗(U).Consequently this solves the problem proposed at the beginning of the lecture.

Proof. It’s easy to see i is a group homomorphism.To see it’s injective, we show 1G 6= g ∈ G gives T g isn’t the identity automorphism. i.e.

we need to find some representation for which g acts nontrivially on V . Now just takethe regular representation, which is faithful!

The hard part is showing that it’s surjective. For this we want to reduce it to theregular representation.

Lemma 6.6

Any T ∈ Aut⊗(U) is completely determined by Tk[G](1k[G]) ∈ k[G].

Proof. Let (V, φ) be a representation of G. Then for all v ∈ V , we have a uniquemorphism of representations

fv : k[G]→ (V, φ) by 1k[G] 7→ v.

If we apply the forgetful functor to this, we have a diagram

k[G] U(V, φ)

k[G] V

Tk[G]

fV

T(V,φ)

fV

1k[G] v

Tk[G](1k[G]) T(v,φ)(v)

Tk[G]

fV

T(V,φ)

fV

�

Next, we claim

22


Lemma 6.7

Tk[G](1k[G]) is a grouplike element of k[G].

Proof. Draw the diagram

k[G] k[G]⊗ k[G] k[G]⊗ k[G]

k[G] k[G]⊗ k[G] k[G]⊗ k[G]

Tk[G]

∆

Tk[G]⊗k[G] Tk[G]⊗k[G]

∆

and note that it implies

∆(Tk[G](1k[G])) = Tk[G](1k[G])⊗ Tk[G](1k[G]). �

This implies surjectivity, by our earlier observation that grouplike elements in k[G] areexactly the elements of G.

Question: does this generalize to e.g. compact Lie groups?

Exercise 6.8. Make Ug into a Hopf algebra with comultiplication

∆(x) = x⊗ 1 + 1⊗ x x ∈ g.

§§6.4 Centers

Definition 6.9. If M is a g-module, then the g-invariants in M are the elements of

Mg = {m ∈M | g ·m = 0 g ∈ g} .

Define the adjoint action as follows: for x ∈ g and a ∈ Ug, we define the action

adx(a) = xa− ax.

Then, we can consider U as a g-algebra with this action.

Proposition 6.10

Z(Ug) = (Ug)g.

Proof. Note C ∈ Z(Ug) is equivalent to Cx = xC for all x ∈ g; equivalently adx(C) = 0for all x ∈ g, meaning C ∈ (Ug)g.

Next time: we prove for “semisimple” Lie algebras, the corresponding representationsare Lie algebras too. We have already seen a counter-example: if g = (C, [, ] = 0) thenthere is a two-dimensional representation

g→ gl(V ) g 7→(

0 g0

)which is not semisimple. The issue “is” upper triangularity.

On the other hand, consider the Lie subalgebra

b =

{(α β0 −α

)| α, β ∈ C

}.

It is a Lie subalgebra of sl2 C, meaning [b, b] ⊆ b] and b is a vector subspace. This is anexample of a solvable Lie algebra:

23


Definition 6.11. A Lie algebra g is solvable if the derived series

D0g = g

Di+1g = [Dig, Dig] i ≥ 0

eventually terminates.

This means that the quotients Di+1/Di in the derived series are abelian Lie algebras.

24



§§7.1 Brief remark on monoidal categories

We already defined a monoidal category: more fully it is a tuple (C ,⊗, a,1, ι) and a mapι : 1⊗ 1→ 1. These have to satisfy some axioms.

Then a tensor category additionally requires C abelian, plus the ⊗ should be “rigid”:every object needs a right dual X∗ giving maps

X∗ ⊗X ev−→ 1coev−−−→ X ⊗X∗

and a left dual ∗X.From this one can deduce that −⊗X has both a left and right adjoint, hence ⊗ has

to be exact.

§§7.2 Ideals

Recall sl(2) has a Lie subalgebra

b =

{(α β0 −α

)| α, β ∈ C

}and we have

b ⊃ [b, b] =

{(0 n0 0

)}def= n ⊃ 0.

We see that n is an ideal of b, in this sense:

Definition 7.1. Let h be a sub Lie algebra of g. We say h is an ideal if [g, h] ⊆ h.Then, g/h makes sense as a Lie algebra.

(This is symmetric, so no distinction between left or right ideals.)We have the “first isomorphism theorem”:

Proposition 7.2

If f : g1 → g2 is a morphism of Lie algebras:

• ker f is an ideal of g1.

• im f is a subalgebra in g2.

• im f = g1/ ker f .

§§7.3 Solvable and nilpotent algebras

Definition 7.3. The Lie algebra g is solvable if the derived series

D0g = g, Dig = [Di−1g, Di−1g]


Definition 7.4. The Lie algebra g is nilpotent if the lower central series

D0g = g, Dig = [g, Di−1g]


25


Lemma 7.5

Nilpotent implies solvable.

Proof. We observe by induction that Dig ⊆ Dig for all i. Thus if Dkg = 0 for k � 0 itfollows that Dkg = 0 for k � 0.

Lemma 7.6

g is solvable if and only if [g, g] is nilpotent.

Proof. We prove only that [g, g] nilpotent implies g solvable. We of course know [g, g] issolvable, and also g/[g, g] is abelian (hence solvable).

Then we appeal to the fact that if we have a short exact sequence

0→ I → g→ g/I → 0

and I, g/I are solvable, so is g.

Definition 7.7. A Lie algebra g is called

• semisimple if there are no nonzero solvable ideals.

• simple if g is not abelian and there are no nontrivial ideals.

Proposition 7.8

g simple implies g semisimple.

Proof. Consider the ideal [g, g] ⊆ g. Since g is simple and not abelian, this forces[g, g] = g.

So g is not a solvable ideal. Since g is also the only ideal by assumption, this implies gis semisimple.

Example 7.9

sl(2,C) is simple. To see this let e, f , h be the usual basis. Then adh = [h,−] :sl(2,C)→ sl(2,C) has the eigenvalues

• 2 with eigenvector e

• −2 with eigenvector f

• 0 with eigenvector h.

In particular, adh(I) is diagonalizable. Then given any ideal I, we have adh(I) ⊆ I,by definition.

Thus, any I needs to be spanned by a subset of {e, f, h} (invariant subspace of adiagonal matrix).

Then we can check that the presence of any of these implies the presence of theother three. (First, [e, f ] = −[f, e] = h so if either e, f ∈ I then h ∈ I. Also, h ∈ Iimplies [h, e] = 2e and [h, f ] = −2f are in I.)

26


Exercise 7.10. Convince yourself this works for sl(2, k) if k is an algebraically closedfield of characteristic other than 2.

On the other hand, show the span of h in sl(2,F2) is a proper ideal. So sl(2,F2) is notsimple.

Example 7.11

Let b = gl(n, k) be the sub algebra of upper triangular matrices and n the strictlyupper triangular matrices. We claim that b is solvable and n is nilpotent.

Definition 7.12. A flag on a finite dimensional vector space V is a sequence

0 = V0 ( V1 ( · · · ( V.

Define for a flag F the submatrices

bF = {x ∈ glV | xVi ⊆ Vi}nF = {x ∈ glV | xVi ⊆ Vi−1}

The standard flag is the obvious flag F std on the standard basis.

Let ak(F ) = {x ∈ glV | xVi ⊆ Vi−k}. Hence b(F ) = a0(F ) and n(F ) = a1(F ).Observe that if x ∈ ak(F ) and y ∈ a`(F ) then the standard product xy lies in ak+`(F ).In particular,

[ak, a`] ⊆ ak+`.

In particular, Din ⊆ ai+1 by induction.Note also that ak(F ) = 0 for all k � 0. In particular, n(F ) is nilpotent, and thus this

proves n(F std) = n is nilpotent.On the other hand, it is not true that b(F ) is solvable for any flag F . However, it is

with the assumption:

Definition 7.13. A flag is complete if it has maximal length (i.e. dimVi+1−dimVi = 1).

We’ll show it just for b(F std).

Claim 7.14. [b, b] = n. In other words, a matrix M is strictly upper triangular if andonly if it can be written as M1M2 −M2M1 for upper triangular matrices M1, M2.

Proof. First, we claim [b, b] ⊆ n. This is because [x, y] ∈ n = a1 for x, y ∈ F std, just bylooking at the diagonal entries of the multiplication (as (xy)ii = (yx)ii).

To show the reverse inclusion, we compute directly with a basis.

Consequently,[b, b] = n = a1

is nilpotent, so b is solvable by the lemma.On the other hand, we would like to show b is not nilpotent. Compute

D1b = [b, b] = n then D2b = [b, D1b] = [b, n]. = n

and hence Dib = n for all i ≥ 2.

27


§§7.4 Representation theory of solvable Lie algebras

We will show that a solvable Lie algebra has only one-dimensional irreps.

Proposition 7.15

Let g be solvable and ρ : g→ gl(V ) a representation. Then there exists a commoneigenvector of ρ(x) across all x ∈ g.

Proof. By induction on dim g. Since g is solvable, we have

g ( [g, g].

Let g′ be any codimension 1 subspace of g containing [g, g]; thus g′ is an ideal (and inparticular a subalgebra); hence solvable.

Then as vector spaces we haveg = g′ ⊕ Cx

for some 0 6= x ∈ g. Now by induction hypothesis there is v ∈ V where ρ(h) = λ(h)v forall v ∈ g′.

Let W denote the span of v0 = v, v1 = ρ(x)v, v2 = ρ(x)ρ(x)v, . . . .

Claim 7.16. This is a g-subrepresentation.

Proof. Just need to check it’s preserved by g′, because preservation under x by construc-tion. We do this by induction on k (with k = 0 being by hypothesis). Then

hvk = hxvk−1 = xhvk−1 + [h, x]vk−1

and [h, x] ∈ g′ since g′ is an ideal.In fact, we actually have

hvk = λ(h)vk + λ([h, k])vk−1. �

To complete the proof of proposition, consider a minimal n such that vn+1 is in thespan of v0, v1, . . . , vn. Then, ρ(h) acting on v0, v1, . . . , vn is an upper triangular matrixwith λ(h) on the diagonal, hence

TrW (ρ(h)) = (n+ 1)ρ(h).

But0 = TrW [ρ(x), ρ(h)] = TrW ρ([x, h]) =⇒ λ[(x, h)] = 0.

Now any w ∈W is a common eigenvector of g′, so we can choose a w ∈W which is aneigenvector for x (which requires the fact that this is a complex representation for aneigenvector to exist).

Corollary 7.17

Any irreducible complex representation of a solvable Lie algebra is one-dimensional.

Thus

• Studying all representations of solvable Lie algebras is hard, but the irreducibleones are all one-dimensional.

• Studying representations of semisimple Lie algebras reduces to the study of justirreps, but the irreps in this case are much more complicated.

28

Evan Chen (Spring 2016) 8 March 1, 2016

§8 March 1, 2016

Last time we saw that if g is solvable, then any representation ρ : g → gl(V ) has acommon eigenvector for all of g. (Hence the irreducibles are one-dimensional, but g mightbe indecomposable.)

§§8.1 Lie’s theorem and Engel’s theorem

Corollary of this:

Theorem 8.1 (Lie’s theorem)

If ρ : g→ gl(V ) is a complex representation of solvable g then there exists a basis ofV such that ρ(x) is upper triangular in this basis (for all x ∈ g).

Proof. There exists a common eigenvector v ∈ V . Consider V/Cv, a quotient representa-tion. Induct down.

Remark 8.2. If g is nilpotent, it’s solvable. (However, we can’t improve this to getzeros on the main diagonal; take the one-dimensional trivial representation.)

Next best thing:

Theorem 8.3

If V is finite-dimensional representation over C and g ⊆ gl(V ) is a Lie algebra suchthat ρ(x) is nilpotent on V for all x, then we can find a basis in which each ρ(x) isstrictly upper triangular.

We now use this to prove:

Definition 8.4. An element x ∈ g is called ad-nilpotent / ad-semisimple if adx :g→ g is nilpotent / semisimple.

Theorem 8.5 (Engel’s theorem)

A Lie algebra g is nilpotent if and only if all x ∈ g are ad-nilpotent.

Proof. If g is nilpotent, then Dng = 0 for some n, which certainly implies (adx)n =[x, [x, [. . . , [x, y]]]] = 0.

Conversely, if adx is always nilpotent, by preceding theorem we can pick a single basisso that adx is strictly upper triangular in every basis. Hence we can build a flag

0 ⊂ g1 ⊂ g2 ⊂ · · · ⊂ gn = g

for which adx gi ⊆ gi−1 for every x, i.Thus each gi is an ideal. We claim Dn−ig ⊆ gi. For i = n it’s clear; inductively if

n− i > m thenDn−i = [g, Dn−i−1g] ⊆ [g, gi+1] ⊆ gi.

Hence Dmg = 0for M � 0, as desired.

29


§§8.2 Jordan decomposition

Definition 8.6. A linear operator A : V → V is

• nilpotent if An = 0 for n� 0.

• semisimple if for all invariant W ⊆ V , there exists an invariant complement.

Let A : V → V , where V is over an algebraically closed field. Then we have thefollowing facts.

(1) The map A is semisimple if and only if A is diagonalizable.

(2) If A is semisimple and W is an A-invariant subspace, then the restrictions of A toW and V/W are semisimple.

(3) The sum of two commuting nilpotent operators is nilpotent.

(4) The sum of two commuting semisimple operators is semisimple.

Theorem 8.7 (Jordan decomposition)

Let V be a vector space over k algebraically closed Any linear operator A : V → Vis uniquely the sum of a nilpotent operator An and a semisimple operator As whichcommute with each other. Moreover, As = p(A) for some polynomial p ∈ k[A], henceAn = A− p(A).

Proof. Generalized eigenspaces. Blah.

Theorem 8.8

Let V be a vector space over k algebraically closed Let A : V → V and define

adA : EndV → EndV adA(B) = AB −BA.

Then(adA)s = adAs

and adAs = P (adA) for some P ∈ k[t], P (0) = 0.

Proof. Let A = As+An, and consider adA = adAs + adAn . It’s easy to see adAs and adAncommute. Choose a basis of V where As is diagonal and As is strictly upper triangular.

Let Eij be the corresponding basis of EndV . Then we have

adAs(Eij) = (λi − λj)Eij

hence adAs is semisimple. On the other hand:

Exercise 8.9 (Kirillov 5.7). adAn is nilpotent.

Hence by uniqueness, we get adas = (adA)s. Moreover, adA a = 0, so P (0) = 0above.

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§§8.3 Jordan decomposition of Lie elements

Definition 8.10. A derivation of an algebra A is any D ∈ gl(A) for which

D(ab) = D(a)b+ aD(b) ∀a, b ∈ A.

The set of derivations is written DerA. It naturally forms a Lie algebra with thecommutator bracket [D1, D2] = D1 ◦D2 −D2 ◦D1.

Hence given a Lie algebra g there is a natural inclusion

g ↪→ Der g x 7→ adx .

A derivation of the form adx is called an inner derivation.

Proposition 8.11

The natural map g ↪→ Der g is an isomorphism when g is semisimple, i.e. allderivations of a semisimple Lie algebra are inner.

Proof omitted, we’ll come back to this.

Theorem 8.12

Let g be a semisimple complex Lie algebra. Then any x ∈ g can be uniquelydecomposed as

x = xs + xn

where xs is ad-semisimple, xn is ad-nilpotent, and xs and xn commute.Moreover, if a y ∈ g commutes with x, it commutes with xs and xn as well.

Proof. First we show uniqueness. We obtain

(adx)s = adxs = adx′s =⇒ adxs−x′s = 0.

So xs − x′s is in the center of g, hence xs − x′s = 0 since g is semisimple.As for existence, decompose g =

⊕gλ into generalized adx-eigenspaces.

Claim 8.13. We have [gλ, gµ] ⊆ gλ+µ.

Proof. To see this, note that for y ∈ gλ, z ∈ gµ, we have

(adx−λid− µid)[y, z] = [(adx−λid)y, z] + [y, (adx−µid)z]

by the Jacobi identity. By induction we eventually get

(adx−λid− µid)n[y, z] =∑k

(n

k

)[(adx−λ)ky, (adx−µ)n−kz]

which implies the result by taking n > dim gλ + dim gµ. �

Now by the (adx)s is a derivation. Hence (adx)s = adxs for some xs ∈ g. Then, set(adx)n = adx−xs .

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§§8.4 Toral subalegbras

Definition 8.14. Let h ⊆ g be a Lie subalgebra. It is called toral if it is commutativeand consists only of semisimple elements in g.

Definition 8.15. A bilinear form 〈−〉 on g is invariant if

〈adx y, z〉+ 〈y, adx z〉 = 0

holds identically for x, y, z.

Theorem 8.16

Let g be a finite dimensional complex semisimple Lie algebra and let h ⊆ g be toral.Let 〈, 〉 be a non-degenerate symmetric invariant bilinear form on g.

1. View g as a h-algebra, and write

g =⊕α∈h∨

gα

where gα are eigenspaces for all h ∈ h with eigenvalues α, meaning

adh(x) = α(h)x h ∈ h, x ∈ gα.

2. [gα, gβ] ⊆ gα+β.

3. If α+ β 6= 0 then gα is orthogonal to gβ with respect to 〈, 〉.

4. For all α, we have a nondegenerate pairing

gα ⊗ g−α → C.

Proof. 1. Since H ∈ h, {adh}h∈h is simultaneously diagonalizable. (We assumed gwas finite dimensional, so gα vanishes for almost all α).

2. Given y ∈ gα, z ∈ gβ we have

adh[y, z] = [adh y, z] + [y, adh z]

= α(h)[y, z] + β(h)[y, z]

= (α+ β)(h)[y, z].

Now [adh y, z] + [y, adh z] = 0 because of g-invariance of the bilinear form.

3. Compute

0 = 〈[h, x], y〉+ 〈x, [h, y]〉= 〈α(h)x, y〉+ 〈x, β(h)y〉= (α+ β)(h) 〈x, y〉 .

4. Similar above.

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From here, we would like to make h as big as possible. It turns out there is “only one”way to do this.

We also need to prove that Der g = g for g semismiple (and in fact the converse istrue).

Finally, we need to show 〈, 〉 actually exists.All of these are related to the existence of a nondegenerate invariant form, whose

existence can be used to detect semisimplicity.

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§9 March 3, 2016

Let H be a closed subgroup of G, where G is a linear algebraic group. Our goal is toshow that G/H has the structure of a variety.

§§9.1 Review of varieties

Recall An = MSpec k[x1, . . . , xn] where k is algebraically closed, with the usual Zariskitopology. Since k[x1, . . . , xn] is Noetherian, ascending chains of ideals stabilize; a “closedset” then corresponds to the vanishing set of some ideal or polynomials.

Let X be an affine variety closed in An = Cn, thus with the Zariski topology. Then wecan consider Xan which is X with the inherited analytic topology of An.

Recall that X is separated if the diagonal morphism ∆ : X → X × X has closedimage im ∆ ⊆ X.

Proposition 9.1 (Properties of Xan)

(i) The inclusion Xan → X is continuous.

(ii) If X → Y is a morphism of varieties, the corresponding map Xan → Yan iscontinuous.

(iii) If X is separated then Xan is Hausdorff.

Lemma 9.2 (An is Noetherian)

Let X ⊆ An be a closed variety. Then

(i) Any family of closed subsets of X contains a minimal one.

(ii) Any descending chain X1 ⊃ X2 ⊃ . . . of closed subsets must eventuallystabilize.

Recall that a space is irreducible if it is not the union of two proper closed sub-spaces.

Lemma 9.3

Let X be a topological space.

(i) A ⊂ X is irreducible if and only if A is.

(ii) If f : X → Y is continuous and X is irreducible, then im f ⊆ Y is irreducibleas well.

Let U be open on X. A k-valued function f on U is called regular at x ∈ U if insome neighborhood x ∈ V ⊆ U ∩D(h) we have f(y) = g(y)h(y)−1 for y ∈ V .

This gives us a sheaf of regular functions on U . Definition of a sheaf deferred toNapkin.

Then, for us we define

Definition 9.4. An abstract variety X consists of a pair (X,OX) where

34


• X is quasi-compact topological space. (any open cover has a finite subcover).

• X is locally affine algebraic and separated.

• OX is a sheaf of rings of k-valued functions on X.

Definition 9.5. • A projective variety is a closed subvariety of Pn.

• A quasi-projective variety is a open subvariety of Pn.

• A locally closed subset is the intersection of an open or closed set.

§§9.2 Algebraic groups

Definition 9.6. An algebraic group is a group that is an algebraic variety, such thatthe multiplication and inversion are regular functions.

For example, GL(n,C) is a distinguished open of An2; hence its an affine variety and

then an algebraic group.

Proposition 9.7 (Connected components algebraic groups)

The irreducible components of an algebraic group G are the connected componentsof G as a topological space.

Hence, we generally think about connected algebraic groups.

Definition 9.8. Any (Zariski) closed subgroup of GL(n,C) is a linear algebraic group.

§§9.3 Group actions

Now suppose G is an algebraic group acting on a space X. Given a point x ∈ X. Then

StabG x = {g | g · x = x} ⊆ G

is the isotropy group, a closed subgroup of G.

Lemma 9.9

Let a group G act on a space X.

(i) G orbits are open in their closures.

(ii) There exists closed orbits.

Proof. We skip (1). Now (2) follows from Noetherian property.

Definition 9.10. We say T is unipotent if T − id is nilpotent.

Proposition 9.11 (Jordan Decomposition)

Let a ∈ GL(V ), then there exists unique as, au ∈ GL(V ) such that as is semisimple,au is unipotent, and

a = asau = auas.

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Proof. Let a = as + an be the Jordan decomposition. Since a is invertible, it has no zeroeigenvalues, thus as is invertible and we can take au = 1 + a−1

s an.

Now, assume X is an affine variety and an algebraic group G act on it. That is, wehave an action

a : G×X → X

which we require to be a map of varieties. Thus this induces a map

a∗ : k[X]→ k[G×X] = k[G]⊗k k[X].

Here k[X] is the coordinate ring (set of regular functions on X). Then given g ∈ G,x ∈ X, and f ∈ k[X], we have

s(g) : k[X]→ k[X] by s(g)f(x) = f(g−1x).

This gives us a representation G→ GL(k[X]), but since k[X] is usually infinite dimen-sional (as a k-vector space: e.g. take k[A1] ∼= k[x]).

Lemma 9.12

Let V be a finite dimensional subspace in k[X].

(1) There exists a finite dimensional W such that V ⊆W and G ·W ⊆W .

(2) G · V ⊆ V if and only if a∗V ⊆ k[G]⊗ V .

Proof. Enough if V is one-dimensional, say spanned by f . Then we have

a∗f =n∑i=1

ui ⊗ fi

for ui ∈ k[G] and fi ∈ k[X]. Then

s(g)f(x) = f(g−1x) =

n∑i=1

ui(g−1)fi(x).

Since ui(g−1) ∈ k don’t depend on x, we see that s(g)f lives in the span of these finitely

many fi’s, for every g. Now we can just let W be the span of s(g)f as g ∈ G, which isfinite dimensional.

Now suppose V is G-stable. Choose a basis of V , say {fi}, and extend it to a basis ofk[X], say {fi} ∪ {bi}. If F ∈ V then

a∗f =∑

ui ⊗ fi +∑

vj ⊗ bj

for some ui, vj ∈ k[G]. Then, applying g gives

s(g)f =∑

ui(g−1)fi +

∑vj(g

−1)bj .

If s(g)V ⊆ V , then by assumption we need the second sum to be zero, meaningvj(g

−1) = 0 for all g ∈ G, i.e. vj = 0. In particular, a∗f =∑ui ⊗ fi + 0, hence

a∗f = k[G]⊗ V .

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Lemma 9.13 (Closed subgroups)

Let H be a closed subgroup in G. Then there exists a finite dimensional rationalrepresentation of G containing a line ` = kv ⊆ H such that h · ` ⊆ ` for every h ∈ H.

Proof. Let G act on itself by left multiplication. Since H is closed, there is an idealI ⊆ k[G] such that every function f ∈ I has f |H = 0.

Let V be a finite dimensional G-stable subspace in k[G] containing a finite set ofgenerators of I. Let W = V ∩ I ⊆ V (hence H ·W ⊂W ) and set n = dimW , then

` = ΛnW ⊆ ΛnV

We claim this W and ` is the desired.Then ` is clearly fixed by H, so we need to show that only the elements of H fix `:

Claim 9.14. If x ∈ G, xW ⊆W ⇐⇒ x` ⊆ `.

Proof. Let (v1, . . . , vn) be a basis of W and extend it to a basis (v1, . . . , vp) of V . Wemay also assume vk+1, . . . , vk+n is a basis of xW for our given x ∈ G (for example k = 0if xW = W , else k > 0).

Then vi1 ∧ · · · ∧ vin are a ΛnV basis (i1 < · · · < in as usual). Let

e = v1 ∧ · · · ∧ vn ∈ `

be the basis element of ` = ΛnW , and

f = vk+1 ∧ · · · ∧ vk+n

be a basis element of x` = ΛnxW . Then x · e is a multiple of f .Of course, if k > 0 then the e and f are linearly independent. �

Hence the proof of the lemma.

Corollary 9.15

There exists a G-homogeneous (transitive G-action) quasi-projective X for G togetherwith a point x ∈ X such that

(1) StabG x = H is closed.

(2) The map Ψ : g 7→ g · x defines a separable morphism G0 → Ψg0.

(3) The fibers Ψ are cosets gH for g ∈ H.

37


§10 March 8, 2016

§§10.1 Remarks

Chevalley’s structure theorem states that a connected algebraic group G over a perfectfield has a unique normal affine algebraic subgroup such that the quotient is an abelianvariety.

Philosophical musing: projective varieties are more often compact or better “moreoften occurring in nature”: when we zoom out, projective varieties get smaller (whereassomething like An gets smaller). We’ve been taught all our lives to look at things locally(“zooming in”, or looking at neighborhoods) when for AG purposes we often want to“zoom out”.

Other philosophical musing: in algebraic geometry, one should usually think aboutclosed sets (which are more natural and well-behaved) than open sets. For example, thestatement of compact (Hausdorff) can be phrased as follows:

Definition 10.1. A closed void is an infinite intersection⋂Cα = ∅.

where each Cα is closed. Then a topological space is compact if any any closed void hasa finite sub-cover.

Exercise 10.2. On Thursday we checked that GL(n) is algebraic. Show that in factGL(n) is affine, meaning we can take a closed embedding

GL(n) ↪→ An.

Exercise 10.3. Fill in some details of the proof that G/H is a variety (below).

§§10.2 Review

Last time we showed that

Proposition 10.4

If H is closed in G and we consider the action of G on k[G], then there exists a finitedimensional G-stable vector space V ⊆ k[G] and a one-dimensional subspace ` ⊆ Vsuch that

{g ∈ G | g · ` = `} = H.

For example, when H = G we can take ` = V =∑

g∈G g, but when H ( G we need somemore work.

We then proved that

Proposition 10.5

There exists a quasi-projective X which is G-homogeneous and x ∈ X with StabG x =H.

Proof. Let π : V \ {0} → P(V ) with V as above, and let G act on imπ. If we letx = π(`) ∈ P(V ), then define X to the be the G-orbit of x. Then we have inclusions

X X Pnopen closed

38


§§10.3 Quotients

Let `(v) denote the line {λ · v | λ ∈ k}.

Definition 10.6. A quotient of G by a closed subset H over k is a pair (Q, a), a ∈ Q,such that

(i) Q is a G-homogeneous space, and `(a) is stable under exactly the elements of H,

(ii) a ∈ Q is a point such that we have the universal property: for every (Y, b) with`(b) stable under H there’s a unique G-equivariant φ : Q→ Y such that φ(a) = b.

Theorem 10.7

A quotient (Q, a) exists and is unique up to G-isomorphism.

For existence, (X,x) as in the earlier proposition works.Now let

G/H = {gH | g ∈ G}

as a set and consider the projection

π : G→ G/H.

If we require π to be an open map, then

Exercise 10.8. Check this defines a topology on G/H.

Now we define O a sheaf of k-valued functions on G/H as follows. Given U ⊆ G/Hopen, we set

O(U) ={f : U → k | f ◦ π is regular on π−1(U)

}.

In other words, O is the pushforward of the sheaf on G.

Exercise 10.9. Check it’s a sheaf.

Now G acts on G/H transitively by left translations (as a set). Consider x ∈ G andthe map

gH·xg7−−→ xgH

is an automorphism of the ringed space (G/H,O). (Check this too.)Now assume (Y, b) as in the universal property. Then we wish to show there exists a

G-morphism of ringed spaces

φ : G/H → Y gH 7→ g · b.

This is

1. Well-defined since StabG `(b) ⊇ H.

2. Continuous by looking at the diagram

G g

G/H Y g · b

π

φ

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3. Morphism of ringed spaces. This follows just because we’re assuming our sheavesour functions: that is we define

OY → f∗OX

by composition; if q : U → k (U ⊆ Y ) then we compose(f−1(U)

f−→ Yq−→ k)∈ OX(f−1(U)) = f∗OX(U).

The uniqueness follows by φa = b.In particular, if (X,x) as in the earlier proposition, there is a unique G-morphism of

ringed spaces

G/Hφ−→ X

such that φH = x. Now φ is an open map and has a continuous inverse, so φ is ahomeomorphism.

Still need to check that the sheaves are isomorphisms. (Use Lie algebra.)

§§10.4 Complete varieties

Definition 10.10. An algebraic variety X is complete if for every Y , the map

X × Y πY−−→ Y

is closed.

(Note that X × Y is not the product of Zariski topologies; in general X × Y has moreclosed sets. Take the diagonal A2 for example.)

In particular, An is not complete for any n ≥ 1. For this example, look at the hyperbolaxy − 1 = 0 in A2. Then the image of the projection onto the y-coordinate i A1 \ {0},which isn’t closed.

Proposition 10.11 (Properties of complete spaces)

Let X be complete.

(1) Any closed S ↪→ X is complete.

(2) If Y is complete, so is X × Y .

(3) If φ : X → Y is a morphism then imφ closed and complete.

(4) If X ⊆ Y is a subvariety then X closed in Y .

(5) If X is irreducible then k[X] = k; the regular functions on X are just scalars.

(6) If X is affine, then X is finite.

Proof. (1) and (2) are easy.(3) requires the fact that if φ : X → Y is a morphism, then imφ is constructible,

meaning it’s the finite union of locally closed sets. (In turn follows from Zariski’s maintheorem.) We also need the fact that if X is separable then the graph of a morphismfrom X is closed. . . .

40


Theorem 10.12 (Projective =⇒ complete)

If X is projective, then X is complete.

§§10.5 Parabolic subgroups

Definition 10.13. Let G be a linear algebraic group. A closed subgroup P of G isparabolic if G/P is complete.

Proposition 10.14

Assume G is connected. Then G contains a proper parabolic subgroup if and only ifG is not solvable.

We will prove this on Thursday.

Theorem 10.15 (Borel’s fixed point theorem)

If G is a connected, solvable, algebraic group acting regularly on a non-empty,complete algebraic variety X over an algebraically closed field k, then there is a Gfixed-point of V .

Proof. Also there exists a closed orbit, say G.x. Then StabG x is parabolic in G, sinceG/ StabG x ∼= G.x is closed and complete (complete follows from X complete, G.x closedin X).

Thus by definition StabG x is parabolic. Since G is solvable, there are no properparabolic subgroups, hence G = StabG x and x is a fixed point.

Proposition 10.16

X over C is complete if and only if Xan is compact.

41


§11 March 10, 2016

Taking a break from algebraic geometry, and talking about Lie algebras attached tocompact Lie groups.

§§11.1 Preliminaries

Recall that a bilinear form B on a vector space V (thus B : V → V ∨) is nondegenerateif kerB = {0}.

Remark 11.1. Check that B is G-invariant if V → V ∨ is also a G-intertwiner.

Fact 11.2. For any V ∈ RepG, and g the associated Lie algebra, we also recall that

V G = V g

whenever G is connected.

Definition 11.3. Let V be a g-module. Then we say a bilinear form B on V is g-invariant if

B(ρ(x)v, w) +B(v, ρ(x)w) = 0

for x ∈ g.

Definition 11.4. We say B is

• positive definite if B(v, v) ≥ 0 with equality only if v = 0.

• negative definite if B(v, v) ≤ 0 with equality only if v = 0.

• (negative) semidefinite if B(v, v) ≤ 0 for all v, but equality might hold in placesother than v = 0.

§§11.2 Reductive Lie algebras

Definition 11.5. A Lie algebra g is reductive if its radical equals its center, i.e.

rad g = z(g)

where the radical rad g is the unique maximal solvable ideal in g.

Claim 11.6. If g is finite-dimensional, then rad g exists, is unique, and contains all thesolvable ideals in g.

Proof. Check that sums of solvable ideals are solvable. It’s easy to see it’s an ideal; tosee solvable, look at the exact sequence

I1 ↪→ I1 + I2 � (I1 + I2)/I1 ' I2/(I1 ∩ I2)

and note the left and right terms are solvable (right is quotient of solvable ideals), hencethe center term.

Then let rad g =∑

I solvable I.

In what follows, assume always g is finite-dimensional.Let (V, ρ) be a representation of g. Now consider the bilinear form on g by

BV (x, y) = TrV (ρ(x)ρ(y)).

We claim that if BV is invariant under the connected group G, it is also invariant underthe action of g. Compute

BV (gxg−1, gyg−1) = TrV (gρ(x)ρ(y)g−1) = TrV (ρ(x)ρ(y)). = BV (x, y)

42


Fact 11.7. If V is an irreducible C-representation, then any h ∈ rad g acts on V by ascalar. Moreover, it acts by zero if h ∈ [g, rad g].

Proof. Since rad g is solvable, there is a common eigenvector v for all h ∈ rad g meaningρ(h) · v = λ(h)v. Hence we get λ : rad g→ C. Then we see that

Vλ = {w ∈ V | ρ(h)w = λ(h)w ∀h ∈ rad g} ⊆ V

is a subrepresentation. Since V is irreducible, Vλ = V .

Proposition 11.8

Suppose BV is non-degenerate for some representation (V, ρ) of g. Then g is reductive.

Proof. We need to show [g, rad g] = 0. Let x ∈ [g, rad g], so adx = 0 for any representationVi, meaning x ∈ kerBVi .

Exercise 11.9 (Kirillov 5.1). Show that if 0 → V1 → W → V2 → 0 is a short exactsequence of g-representations then BW = BV1 +BV2 .

Then by induction, we get x ∈ kerBV , but BV was nondegenerate hence x = 0

Theorem 11.10

gl(n,C) is reductive.

Proof. Let glnC act on C⊕n = V in the usual way. Then

BV (x, y) =∑

xijyji

is nondegenerate as desired.

We havegln = z(gln)︸︷︷︸

=C

⊕ sln(C).

Now note that [gln, gln] = sln, and sln is reductive for the same reason as above: withrespect to BV , z(gln) and sln are orthogonal.

§§11.3 Killing form

View g as a representation over itself by ad, i.e.

ad : g→ gl(g)

Definition 11.11. The Killing form is the bilinear form Kgdef= Bad(x, y) = Tr(adx ady).

Remark 11.12. Note that:

z(g) = ker ρ ↪→ g→ gl(g).

Theorem 11.13 (Cartan’s criterion)

The Lie algebra g is semisimple if and only if Kg is nondegenerate.

43


§§11.4 Compact Lie groups

Here is an example which will be an important ingredient in the following theorem (butis not an example of the theorem itself)

Example 11.14

Let g = U(n) be the skew-Hermitian matrices. Then the trace form 〈x, y〉 = Tr(x, y)is negative definite. To see this note, that

Tr(xy) = −Tr(xy†) =⇒ Tr(x2) = −Tr(xx†) = −∑i,j

|xij |2.

Thus g is reductive.

We’ll also need

Theorem 11.15

Let G be a compact real Lie group. Then g is reductive, and the Killing form isnegative semidefinite. Moreover, kerK = z(g) and K is negative definite on thesemisimple part g/z(g) = gss.

Conversely, if g is a semisimple real Lie algebra with negative definite Killing form,then g is attached to some compact Lie group.

Proof. Since G compact, then every complex representation is unitary (first week ofclass). Thus ρ(G) ⊆ U(V ), and ρ(g) ⊆ u(V ). Thus the trace form BV (x, y) is negativesemidefinite with kerBV = ker ρ.

Now apply this to V = gC with G acting on gC by adjoint action. Then BV is theKilling from K, thus negative semidefinite.

Now it’s a theorem that if g is reductive then g = z(g)⊕ gss. But [gss, gss] = gss. ThusKerK = z(g).

For the converse, assume g is a real semisimple Lie algebra with negative definite killingform K. By Lie’s third theorem, let G be a connected Lie group with Lie algebra g,and define positive definite B(x, y) = −K(x, y). Now B is AdG invariant, which meansAdG ⊆ SO(G); since AdG is connected it follows AdG is the connected component ofthe identity of Aut g (the group of Lie algebra automorphisms).

Regard Aut g ↪→ GL g as a closed Lie subgroup. Then im AdG is closed in the compactSO(g) and hence AdG is compact. So

AdG = G/Z(G)

whence the Lie algebra attached to AdG is g/z(g) ' g (the center is trivial since Knondegenerate).

§§11.5 One-dimensional representations

Proposition 11.16

If g is semisimple, then the only 1-dimensional representation is the zero one.

Proof. If ρ : g→ C has [g, g] = g then notice that ρ[g, g] = 0.

44


Proposition 11.17

If g is reductive, then one-dimensional representations of g correspond to one-dimensional representations of z(g).

Contrast this with the solvable case when all irreducible representations were one-dimensional.

We’d like to find some middle ground here. Suppose b ↪→ g, where b is solvable and gis a reductive Lie algebra. Then we get a map z(g) ↪→ b.

Not all irreps of b come from z(g) in this way (for example, if g is semisimple then ghas no center at all). So we want to understand how to uniformly construct all irreduciblerepresentations of b.

Given an irrep V of b, let V be indgb V . It’s not clear V should be irreducible. This is

not always true, but

• If g is semisimple over C, then it’s true.

• In the modular case, it’s not true.

I’m hungry.

§§11.6 Borel and parabolic subgroups

Let V be a representation of b. Let G be semisimple and reductive. Assume b = LieB,where B is a Borel subgroup in G (defined later). We saw that if P is a parabolicsubgroup of G, then G/P was complete and quasi-projective. Then a minimal parabolicP (containing no other proper parabolic) is the Borel subgroup; it keeps as muchG-information as possible. In fact,

Theorem 11.18 (Borel ⇐⇒ Maximal solvable)

Let B be a subgroup of a semisimple, reductive G. Then the following are equivalent:

• B is Borel.

• B is the maximal solvable subgroup of G

• B is the smallest subgroup for which G/B is projective.

• G/B is complete and B is solvable.

Theorem 11.19 (Levi’s theorem)

Given a general Lie algebra g, we have

g = rad g⊕ gss.

the Lie algebra gss is not semisimple.

Remark 11.20. A complex variety is complete if and only if it is compact as a complex-analytic variety.

New office hours: W 11:30AM - 1PM.

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§12 March 15, 2016

Homework 4 due Thursday. Next week: holiday.Today: diagonalizable groups.

§§12.1 Proof of proposition from last time

Last time:

Proposition 12.1

A connected linear algebraic group G contains a proper parabolic subgroup if andonly if G is not solvable.

Proof. Assume G is closed in GL(V ) and let G act on P(V ). Then there exists a closedorbit X, thus X is projective (hence complete).

Let x ∈ X, and P = StabG x; this gives a morphism

G/P → X gP 7→ g · x.

This is bijective. By some lemma, this implies P is parabolic. (Namely, if X and Y areG-homogeneous and φ : X → Y is a bijective G-map of varieties then X complete ⇐⇒Y complete.)

If P 6= G then done. If P = G, set V1 = V/x, and let G act on PV1. Continue in thisfashion. Eventually we’ll get either a proper parabolic, or G will be isomorphic to uppertriangular matrices in GL(n) whence G is solvable.

Conversely, assume G is connected and solvable. Assume P ( G is parabolic and ofmaximal dimension. (Lemma: P parabolic in G =⇒ P 0 parabolic in G0.) Thus we canassume P is connected.

Consider (G,G) the group generated by commutators; this is closed and connected.Let Q be the orbit of P on (G,G); this is connected, parabolic, and contains P . Then

• If Q = G, then (G,G)/(G,G) ∩ P → G/P is a bijection, hence (G,G) ∩ P isparabolic in (G,G). If (G,G) 6= G then induct down; now we may as well assume(G,G) ∩ P = (G,G), or (G,G) ⊆ P , contradiction.

• Else if Q = P , then (G,G) ⊆ P , so G/P is affine.

Here’s a lemma that might be useful:

Lemma 12.2

Let H be a closed subgroup of the linear algebraic group G. Assume π : G→ G/Hhas “local sections”, and H acts on X. Then the fiber product

G×H X ={

(g, x) ∈ G×X | (gh, h−1x) ∼ (g, x)}

exists. This is a fiber bundle for G/H.

Proof: exercise.

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§§12.2 Rational representations

We want to study RepG or Rep g, where RepG means “finite-dimensional rationalrepresentations” in a context where G is an algebraic group.

Definition 12.3. Let G be an algebraic group. Then a representation is a map φ : G→GL(V ) which is a morphism of varieties (where we also view GL(V ) as a linear algebraicgroup).

If we have subalgebra h in g built out of semisimple commuting things (i.e. h is toral)then restricting to h gives a decomposition

g =⊕α

gα.

OK now consider the following.

• The group G acts on itself by conjugation.

• This induces an action Ad of G on g.

• This induces an action ad of g on g.

The way to go from the last of these back to the original (for simply connected Lie groupsG) is through the exp map, which is not algebraic.

Sometimes we want to view G acting on k[G] as a “rational” representation, but wedon’t have finite-dimensional.

Definition 12.4. let V be any vector space (possibly infinite dimensional). Thena ∈ EndV is locally finite if we can decompose V =

⋃α Vα such that each Vα is a-stable

and dimVα <∞.Assuming a is locally finite, we say it is

• semisimple if restriction to any finite dimensional a-stable subspace is semisimple.

• locally nilpotent if restriction to any finite dimensional a-stable subspace isnilpotent.

• locally unipotent if restriction to any finite dimensional a-stable subspace isunipotent.

Then as before, any locally finite a can be written as

a = as + an

where as is semisimple and an is locally nilpotent.If a is invertible and locally finite, we can also put

a = asau

where as is semisimple, and au is locally unipotent.

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Example 12.5

Let Gm be the multiplicative group of the field

Gm = Spec k[x, x−1]

which as a set of points is k∗; this is also GL(1, k).We have Hom(Gm,Gm) ∼= Z since the only algebraic maps from Gm to itself are

the maps x 7→ xn.This gives the irreducible one-dimensional representations, since we have computed

all maps Gm → GL(1, k).On the other hand, since Gm is one-dimensional as a group, so all the irreducibles

are one-dimensional.

Consider G acting on k[G], call this ρ. One can show that ρ(g) is locally finite for allG, so we get a decomposition

ρ(g) = ρ(g)sρ(g)u.

Theorem 12.6

For any g ∈ G, there exist unique commuting gs, gu ∈ G such that ρ(g)s = ρ(gs) andρ(g)u = ρ(gu). Thus homomorphisms of algebraic groups preserve the properties“unipotent” and “semisimple”.

Definition 12.7. Let σ ∈ AutG. We say σ is semisimple if σ looks like conjugationby a semisimple element in GLV .

That is, if we take our embedding φ : G→ H ⊆ GL(n) with H a closed subgroup ofGL(n), then there should be s ∈ GL(n), in the normalizer of H with respect to GL(n),so that

φ(σx) = sφ(x)s−1 ∀x ∈ G.

Exercise 12.8. Let Gu be the set of unipotent elements in G. Show that Gu is alwaysa closed subgroup of G.

§§12.3 Diagonalizable groups

Definition 12.9. We say G is diagonalizable if H is isomorphic to a closed subgroupDn of diagonal matrices in GL(n). We say H is a torus if H ' Dn.

Theorem 12.10

If G is a commutative linear algebraic group then Gs and Gu are closed subgroupsand

π : Gs ×Gu → G

is an isomorphism of algebraic groups.

Reading: Dedekind’s theorem. (Prime ideals in a Dedekind domain.)Now let G be a linear algebraic group, and let χ : G → Gm a homomorphism of

algebraic groups. Then we call χ the rational characters, the set of which is

X∗(G) = Hom(G,Gm) ⊆ k[G]

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which is an abelian group; we’ll often write the operation additively. (Tangential remark:the dual concept is X∗(G) = {Gm → G}; for G commutative this is also an abelian group.If G is arbitrary, this is still a Z-module.)

Dedekind’s theorem says these characters are linearly independent in k[G]. Here is thecondition for it to be a basis:

Theorem 12.11

Let G be a linear algebraic group. The following are equivalent.

(1) G is diagonalizable.

(2) X∗(G) is a finitely generated abelian group and a k-basis for k[G].

(3) RepG is a semisimple abelian category whose irreducibles are one-dimensional.

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§13 March 17, 2016

Last time we saw X∗(G) forms a k-basis of k[G].

§§13.1 More on diagonalizable groups

Definition 13.1. Let M be an abelian group. Then k[M ] is the algebra with basis e(m)for m ∈M , with e(m)e(n) = e(m+ n).

We can make it into a Hopf algebra via

∆(e(m)) = e(m)⊗ e(m)

i(e(m)) = e(−m)

ε(e(m)) = 1.

In particular

Corollary 13.2

if G is diagonalizable, then X∗(G) is a finitely generated abelian group. It has nop-torsion if char k = p > 0.

Moreover, k[G] ∼= k[X∗(G)] always.

Proof. If char k = p > 0 then the only pth root of 1 is 1; hence if X∗(G) contains p-torsionthis contradicts the assumption that X∗(G) forms a k-basis.

To show the isomorphism, we first note:

Claim 13.3. If M1 and M2 are both finitely generated then

k[M1 ⊕M2] ' k[M1]⊗k k[M2].

Claim 13.4. If k[M ] is affine, then there exists a diagonal algebraic group G (M) withk[G (M)] = k[M ] such that ∆, ε, i are comultiplication, inversion, and the unit.

Moreover, there is a canonical isomorphism

M ' X∗(G (U)).

If G is diagonalzible, we also have a canonical isomorphism

G (X∗(G)) ' G.

Proof. We can reduce to the case that M is cyclic, due to the result on k[M1 ⊕M2] =k[M1] ⊗ k[M2]. If M = Z, then k[M ] ∼= k[T, T−1] which is an integral domain. IfM = Z/d with p - d, then k[M ] ' k[T ]/(T d − 1). The polynomial T d − 1 doesn’t havemultiple roots since d - p, so we get a reduced ring, hence affine variety. Now we justcheck the coalgebra properties manually.

Now for m ∈M , the map is

m 7→ [x 7→ e(m)(x)]

where we regard e(m) ∈ k[G (M)] b the isomorphism. This is an isomorphism becauseDedekind’s theorem implies these characters form a basis.

The third part is a similar construction. �

This now completes the proof of the proposition.

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We think of G as an “inverse” to X∗ for G diagonalizable. This gives you:

Exercise 13.5. Fix k algebraically closed. The category of diagonalizable linear algebraicgroups is equivalent to the opposite category of finitely generated abelian groups, via theG and X∗ maps.

Corollary 13.6

Let G be diagonalizable. Then

(i) G is a direct product of a torus and finite abelian group with order prime top = char k.

(ii) G is a torus if and only if G is connected.

(iii) G is a torus if and only if X∗(G) is free abelian.

Proof. First we check that G (Zn) is isomorphic to the diagonal matrices in k. Now weknow X∗(G) = Zn ⊕M for some finite abelian group M , hence G ' Dn × G (M).

§§13.2 Rigidity

Theorem 13.7

Assume G and H are diagonalizable. Let V be a connected affine algebraic variety.Assume φ : V ×G→ H is a morphism of varieties such that for all v ∈ V , the mapφ(v,−) : G→ H is a morphism of algebraic groups.

Then for x ∈ G, the function φ(−, x) is constant.

Proof. Fix v ∈ V and let ψ ∈ X∗(H). Then

ψ(φ(v,−)) ∈ X∗(G)

by composing

Gφ(v,−)−−−−→ H

ψ−→ Gm.

By taking our basis, we have

ψ(φ(v,−)) =∑

x∈X∗(G)

fx,ψ(v)︸︷︷︸∈k[V ]

χ(x).

By Dedekind’s theorem, for every v, we see fx,χ(v) is 1 for some x and 0 elsewhere. Thusf2x,χ = fx,χ. Since V is connected we get fx,χ is 1 for some x and 0 elsewhere.

Application: if G is any linear algebraic group (not necessarily diagonalizable) and His closed in G, then ZG(H) is closed in NG(H), and NG(H)0 = Z(H)0; in particular thequotient N(H)/ZG(H) is finite.

Fact 13.8. If G is a linear algebraic group then G0 is normal in G and |G/G0| <∞.

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§§13.3 Tori

§§13.4 Parabolic subgroups

Lemma 13.9 (Parabolic is transitive)

Let P be parabolic in G, and Q be parabolic in P . Then Q is parabolic in G.

Sketch of proof. We’d like G/Q to be complete. We need for all X that πX : G/Q×X →X to be closed. The idea is to rewrite πX as the composition of closed maps (usingcompleteness of G/P , P/Q).

Assume this for now. Then:

Lemma 13.10

(i) Let P be parabolic in G. If Q is closed in G and contains P , then Q is itselfparabolic.

(ii) P is parabolic in G if and only if P 0 is parabolic in G0.

Proof. (i) We have a surjective morphism G/P � G/Q. The image from a completespace is complete.

(ii) P is parabolic in G while P 0 is parabolic in P . Also G0 parabolic in G, . . .

§§13.5 Borel subgroups

Definition 13.11. A Borel subgroup of G is

• closed, connected, solvable, and

• maximal with respect to the above properties.

Theorem 13.12

(1) A closed subgroup of G is parabolic if and only if it contains a Borel subgroup.

(2) A Borel subgroup is parabolic.

(3) Two Borel groups in G are conjugate.

Proposition 13.13

Let φ : G� G’ be a surjective morphism of linear algebraic groups.

(i) If P is parabolic in G, then φ(P ) is parabolic in G′.

(ii) If B is Borel in G, then φ(B) is Borel in G′.

52


§14 March 29, 2016

Guest lecture by David Vogan.

§§14.1 Exercises for Tuesday April 5

Exercise 14.1. Let G be a linear algebraic group, and H a closed subgroup. Prove thefollowing are equivalent.

(a) G/H is an irreducible algebraic variety.

(b) H meets each connected component of G.

(c) G = G0.

Exercise 14.2. Give an example of a finite solvable

H ⊆ SL(2,C)

so that H is not conjugate to a group of upper triangular matrices.

Exercise 14.3. Let G be a connected linear algebraic group, all elements are semisimple.Prove that G is a torus. (Consider a Borel subgroup of G.)

§§14.2 Matrices

Let

g =

a1 x12 . . . x1n

a2 . . . x2n

. . ....an

.

Clearly, g is unipotent iff all ai are equal to 1.Now consider semisimple. We know that:

• If all xij = 0 then g is semisimple.

• If all ai are distinct, then g is semisimple.

On the other hand, for example let n = 2. Then g is semisimple if and only if one of theabove hold.

In fact, in generalg semisimple ⇐⇒ xij = 0∀ai = aj .

For example, we can explicitly write n = 2 as(a1 x12

0 a2

)is semisimple ⇐⇒ a1 6= a2 or x12 = 0.

The interesting thing to note is that for the latter case, this condition is actually neitheropen nor closed : instead it is the union of an open set and a closed set.

Conclusion: if G is linear algebraic then we have Gu ⊆ G the set of unipotent elements,which is a closed subset of G (not necessarily subgroup). On the other hand the set ofsemisimple elements Gs usually neither open nor closed.

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§§14.3 Lie-Kolchin

Theorem 14.4 (Lie-Kolchin)

If G ⊆ GL(n, k) is a (closed) connected solvable subgroup then there exists x ∈GL(n, k) so that

xGx−1 ⊆ Tn = {upper triangular} .

(Hence the second exercise earlier says that we need the “connected” assumption.)

Proof. Induction on n. Let GL(n, k) act on projective space Pn−1. The Borel fixed pointtheorem implies G has a fixed point k · v: thus v is an eigenvector across all of G. Modout by v and induct down.

Remark 14.5. Wikipedia: “the Borel fixed-point theorem is a fixed-point theorem inalgebraic geometry generalizing the Lie-Kolchin theorem.”

Thus, the right way to picture a (connected) solvable subgroup is just as a subgroupof upper triangular matrix (i.e. these are the only examples, up to conjugation).

Lemma 14.6

Let V be an n-dimensional vector space over an algebraically closed field k, and let Sbe a commuting set of diagonalizable endomorphisms. Then S is simultaneouslydiagonalizable.

Proof. Induction on n. If all elements of S are scalars, result is immediate.Thus assume s1 ∈ S is not a scalar, so it has an eigenvalue λ1 but isn’t λ1 · id. Let V1

be the λ1 eigenspace, and V2 the sum of all other eigenspaces of S1. Thus we can write

V = V1 ⊕ V2

with dimV1,dimV2 < n. Thus if s2 ∈ S commutes with s1 then it preserves thedecomposition. So induct downwards.

Theorem 14.7

Let G be a connected abelian linear algebraic group. Define Gs as the set ofsemisimiple elements and Gu as the set of unipotent elements. Then Gs and Gu areclosed subgroups, and the group multiplication gives an isomorphism

Gs ×Gu → G

of algebraic groups.

Proof. Assume G ⊆ GL(n, k).Now, Gs and Gu closed under multiplication since G is commutative, so they are

abstract groups.Recall Gu is a closed algebraic subgroup. Since Gs is a commuting family of diagonal-

izable matrices, we can simultaneously diagonalize them (previous lemma).

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Let Dn be the diagonal matrices of the ambient space in this basis. It consists ofsemisimple matrices and containts Gs, so Gs = G ∩Dn. This completes the proof thatGs and Gu are Zariski closed.

Now Gs∩Gu = {1}, and so the morphism Gs×Gu → G is injective. It is also surjectiveby Jordan decomposition. So it’s a bijective morphism of algebraic groups. Omit proofthat it’s actually an isomorphism (obvious in characteristic 0, more work in p).

Corollary 14.8

Retain the setting of the previous theorem. Then Gs and Gu are connected.

Proposition 14.9 (Connected linear algebraic group of dimension 1)

If G is connected linear algebraic group of dimension 1, then it is commutative andeither G = Gs ∼= k× = Gm or G = Gu ∼= k = Ga.

Proof. Fix g ∈ G. Define ϕ : G→ G by

ϕ(x) = xgx−1

hence imϕ is a conjugacy class of g ∈ G.Since G is connected it’s irreducible as a variety. Thus closure of imϕ is irreducible

and connected in G. But since G is one-dimensional we either have

(i) imϕ is 0-dimensional, which means that g is in the center Z(G).

(ii) The closure im(ϕ) equals G. Thus almost everything in G is conjugate to g.

We expect a postieri the second case never occurs, so let’s dispense of it. By hypothesisthe characteristic polynomial is constant on conjugacy classes, hence on G. In particular,everything in G has the same characteristic polynomial as 1, hence in particular isunipotent. By Engel’s theorem this implies that we can choose a basis in which it isupper triangular with all entries 1. Then the commutator [G,G] is a proper subgroup ofG, hence of lower dimension, so [G,G] = 0, which contradicts the assumption we’re inthe second case.

Thus g ∈ Z(G) for every g ∈ G, and we know G is commutative. As before we knowhave an isomorphism

Gs ×Gu → G

and again because G has dimension 1, we either have G = Gs or G = Gu.So now we have to show that

• A commutative connected semisimple algebraic group is copies of Gm. (Easy, maybedo on Thursday.)

• A commutative connected unipotent algebraic group is copies of Ga.

Remark 14.10. Look at SL(2, k) and sl(2, k) where k has characteristic 2. Then SL(2, k)is not solvable, but sl(2, k) is.

55


Remark 14.11. Look at

T 2 =

{(a b0 a−1

)}again where k has characteristic 2. This is non-abelian and solvable. But t2 has basis(

1 00 1

),

(0 10 0

)which is abelian.

Consider the following (will discuss Thursday):

Exercise 14.12. Let g =

(a 00 b

)and consider 〈g〉 ⊆ Dn. What is its Zariski closure?

On Thursday: will prove existence and conjugacy of maximal tori.

56


§15 March 31, 2016

§§15.1 Maximal Torus

Recall:

Definition 15.1. A torus is a connected abelian linear algebraic group consisting ofsemisimple elements.

Hence it’s isomorphic to (k×)m.Given G a linear algebraic group, a maximal torus is a subtorus of maximal dimension.The main theorem is

Theorem 15.2

Let G be a connected solvable linear algebraic group. Then

• Any semisimple element s of G belongs to some maximal torus.

• The centralizer ZG(s) is connected.

• Any two maximal tori in G are conjugate by the lower central series D∞G,the intersection of Di(G) = [G,Di−1(G)].

Finally if T ⊆ G is a maximal torus, then G ∼= T oGu; any g ∈ G can be uniquelywritten as t · u.

Remarks:

Corollary 15.3

Maximal tori in any connected linear algebraic group G are all conjugate to oneanother.

Proof. If T and T ′ are maximal in tori, extend them to maximal solvable subgroups B,B′ (Borel subgroups). By conjugacy of Borel subgroups, gBg−1 = B′. Thus gTg−1, T ′

are two maximal tori in B′, thus conjugate in B′.

§§15.2 Main idea

The argument for our main theorem will differ slightly from Springer (following Humphreys’“Linear Algebraic Groups”).

Let G be solvable. Thus we can think of it as a subset of the upper triangular matricesTn for some n:

G ⊆ Tn =

∗ . . . ∗

0. . .

...0 0 ∗

We know in that case Gu = G ∩ Un, a closed connected normal subgroup of G; here Unis the unipotent elements.

Let Dn be the diagonal subgroups now. Of course we have an exact sequence

1→ Un ↪→ Tn � Dn → 1.

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Thus there is an exact sequence

1→ Gu ↪→ G� S → 1

where S is the image of G in Dn (set of diagonal matrices). The projection above is amorphism of algebraic groups, hence it has a connected images. Thus S is a connectedsubgroup of Dn, meaning it is a torus of some dimension ≤ n.

Consequently, if G is solvable then G/Gu is a torus S. Moreover, if T is any subtorusof G, then T maps injective into a subtorus of G/Gu, hence dimT ≤ dimS.

The main content is actually finding a subtrous T ⊆ G such that dimT = dimS. Todo this, we use induction on dimG.

Let’s first consider the case where G is nilpotent, meaning that for some N > 0 wehave DN (G) = 1.

Proposition 15.4

A linear algebraic group G is nilpotent if and only if Gs is a subgroup of Z(G)

Proof. Assume G nilpotent, s ∈ Gs. Consider the map

χ : G→ G by x 7→ sxs−1x−1

Since G is nilpotent, χN (x) = 1 for N large; But χ is a morphism of algebraic varieties,and the differential dχ is Ads−id.

Since s is semisimple, Ads is semisimple. So dχ is semisimple as well. But it’s alsonilpotent as (dχ)N = 0, thus dχ = 0. Hence Ads = id, which means s ∈ Z(G).

Proposition 15.5

If G is nilpotent and connected, then

(i) Gs, Gu are closed subgroups.

(ii) Gs is a central torus.

(iii) G ∼= Gs ×Gu.

We proved this last time for commutative groups G. We can repeat this proof frombefore using just using the fact that G is central.

Corollary 15.6

If G is connected and solvable, then

(i) [G,G] is closed, connected, unipotent, and normal.

(ii) Gu is closed, connected, unipotent, normal.

(iii) G/Gu is a torus.

(We more or less did this earlier.)

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§§15.3 Proof of main theorem

If Gs is central in G, then G is nilpotent, and we’re done by earlier work. (Here Gs isthe unique maximal torus.)

Otherwise, pick s ∈ Gs which is not in Z(G). So Ads is diagonalizable, but not zero.Let H denote the centralizer of s in G, which is a (possibly disconnected) linear solvablealgebraic. The corresponding Lie algebra h is the zero eigenspace of Ads which is a propersubalgebra of the Lie algebra g of G. Consequently,

dimH < dimG.

Consider G ⊆ Tn and SG the image described earlier. By induction, H has a maximaltorus TH ∼= SH . The aim is to show that SG = SH . We won’t be able to prove thisproperly in this class.

§§15.4 An example!

Let

G = T2 =

{(a c0 b

)}.

Then

Gu = U2 ==

{(1 c0 1

)}.

The maximal tori are elements of the form(1 x0 1

)(a 00 b

)(1 −x0 1

)=

(a (b− a)x0 b

).

We call the set of these Tx (fix x, varying a, b).The semisimple elements are (

a y0 b

)such that a 6= b or y = 0. If y = 0 then it belongs to every maximal torus, otherwise ify 6= 0 (hence a 6= b) it belongs to the maximal torus corresponding to

x =y

b− a.

§§15.5 Cartan subgroups

Assuming Theorem 10, it follows that maximal tori in any connected linear algebraicgroup are all conjugate to each other.

Definition 15.7. A Cartan subgroup of a connected linear algebraic group is theidentity component of the centralizer of some maximal torus (which are all conjugate).

These are “not important”: used by Cartan in structure theory of complex Lie algebrasbefore there were algebraic groups. These days it is mostly a “historical curiosity”.

§§15.6 Radicals

If G is connected, then define the nilpotent radical of G, Ru(G), to be the largestunipotent normal subgroup. This is the intersection of unipotent radicals of all the Borelsubgroups of G.

The radical of G is the largest solvable normal subgroup, which is the intersection ofall Borel subgroups. Of course R(G) ⊇ Ru(G).

We say G is semisimple if R(G) = 1 and reductive if Ru(G) = 1.

59

Evan Chen (Spring 2016) 16 April 5, 2016

§16 April 5, 2016

To do: conjugacy and existence of maximal tori in connected, solvable groups.

§§16.1 Review

Quick review of connected solvable groups B. We can assume B is a closed subgroup ofTn by Lie-Kolchin.

Given connected commutative algebraic group C, we have C = Cs×Cu (isomorphism),where Cs and Cu are closed connected subgroups. Given connected nilpotent algebraicgroup N , we have N = Ns ×Nu where Ns is central torus and Nu is normal in N .

But given connected solvable algebraic group B, we would like Bs ×Bu → B to be anisomorphism. But Bs is too big to be a diagonal subgroup.

Note that if T is a torus then ZG(T ) contains lots of semisimple elements. However,we want to prove that

Theorem 16.1

The maximal torus T gives an isomorphism of varieties

T ×Bu → B

where T is no longer central, but Bu is normal in B.

Note that Z(B)s ⊆ T , because z ∈ Z(B)s lies in a maximal torus, and all maximaltori are conjugate to one another. But the central elements are stable under conjugation.

§§16.2 More on maximal tori

Corollaries of this theorem:

Corollary 16.2

Let G be a connected solvable linear algebraic group. Suppose H ⊆ G is a closedsubgroup consisting only of semisimple elements. Then

(1) H lives in a maximal torus.

(2) ZG(H) is connected and equal to NG(H).

Proof. First, let’s see H is commutative We have a diagram

G G/Gu

H H/Hu

where G/Gu is a torus. So H is isomorphic to a subgroup of a torus, hence commutative.If H ⊆ Z(G), we’re okay. Otherwise, if s ∈ H with s not central, we know ZG(s) is

connected.So H ⊆ ZG(s), the latter being solvable (since it’s a subgroup of G).

60


Use induction on dimG; we simply need dimZG(s) < dimG. As s /∈ Z(G), we getZG(s) ( G. Thus ZG(s) is a proper, closed, connected irreducible subgroup of G.

The fact that ZG(H) is connected follows from the same induction as above. For thesecond part, if x ∈ NG(H) and h ∈ H, then xhx−1h−1 ∈ H ∩ (G,G) ⊆ H ∩ Gu = {1}since H was assumed to consist of semisimple elements. Therefore x ∈ ZGH) too.

§§16.3 Maximal tori are conjugate

From now on G is connected, linear algebraic group, but we don’t have hypotheses like“solvable”. We prove anyways that

Theorem 16.3

Maximal tori are conjugate.

Proof. Fix a maximal torus T in G. Fix a Borel B in G and note B has a maximal torusT ′. Since T is connected, solvable, it’s contained in some Borel B′. But B′ is conjugateto B, say

B′ = gBg−1.

Thus T is conjugate to a maximal torus in gBg−1 which is conjugate to T ′.

Proposition 16.4

Let T be a maximal torus in G, and let C = ZG(T )◦.

(1) C is nilpotent, and T is a maximal torus in C.

(2) ∃t ∈ T lying in only finitely many conjugacy classes of C.

(Here C is an example of a Cartan subgroup. We’ll show in fact that ZG(T ) was connectedanyways.)

Proof. Observe T ⊆ Z(C), since the torus T commutes with all of ZG(T ), hence with C.Now let B be a Borel subgroup of C with T ⊆ B ⊆ C. Then also, T ⊆ Z(B), so T is

normal in B. Thus B nilpotent, and B/T ∼= Bu. We now show that

Lemma 16.5

If G is a linear algebraic group and its Borel B is nilpotent, then G0 = B. Moreover,Z(G)0 ⊆ Z(B) ⊆ Z(G).

Proof of Lemma. For the inclusion, Z(G)◦ is closed, connected, commutative, so Z(G)◦

lives in a Borel. Since they’re all conjugate, it lives in all of them, in particular it lives inB.

Also, let g ∈ Z(B). We have a map G→ G by

x 7→ gxg−1x−1 for g ∈ Z(B)

which vanishes on B, so it gives a map G/B → G. Since G/B is projective, G is affine, itfollows it is the constant map. Thus it is trivial, so gx = xg and finally g ∈ Z(B); thusZ(B) ⊆ Z(G).

61


B is nilpotent solvable and a connected nilpotent group, so it contains a nontrivial,closed connected group H in its center. We can take the subgroup generated by maximallength commutators. . . . � fill infill in

Exercise 16.6. If φ : G → G′ is a homomorphism of algebraic groups, then ϕ(G◦) =ϕ(G)◦.

Thus C = B, and C is nilpotent.

Lemma 16.7

If S is a subtorus of G, then there exists s ∈ S such that ZG(s) = ZG(S).

Proof. We can assume G = GLn and S ∼= Dn. Let χ1, . . . , χm be distinct characters ofχ. Thus there exists an s such that χi(s) 6= χj(s). We claim that for such an s, we haveZG(s) = ZG(S). IT suffices to check ZG(s) ⊆ ZG(S), but if q ∈ ZG(s) then q must fixall of χ1, . . . , χm which is a basis for the character group of S.

Remark 16.8. The set of such s are dense, open in S.

We can now prove the part of the proposition: let t ∈ T with ZG(t) = ZG(T ). Ift ∈ gCg−1, then g−1tg si semisimple, and lives in C = ZG(T )◦. Thus g−1tg ∈ T .

Check that ZG(g−1tg) = g−1ZG(t)g ⊇ T . So g ∈ NG(t).

Lemma 16.9

H is a closed subgroup of connected G. Consider the subvariety X =⋃x∈G xHx

−1

of G. Then

(1) X contains nonempty open subsets of X. Moreover, if H is parabolic, then Xis closed.

(2) Assume [NG(H) : H] < ∞. Moreover, there exists elements of H in onlyfinitely many conjugates of H. Then X = G.

The proof of this is mostly algebraic varieties. Proof later, possibly.Now we prove the big theorem.

Theorem 16.10 (1) Every element of G lives in a Borel subgroup.

(2) Every semisimple element in G lives in a maximal torus.

(3) The union of the Cartan subgroups of G contains a dense open subset.

The last thing is what we’ll use to prove the Cartan subgroup is all of the centralizer.

Proof. Let’s assume T is a maximal torus, and let C denote its Cartan, meaning T ⊆C = ZG(T )◦. Let B be a Borel containing C (this exists because C is nilpotent andconnected).

Apply the lemma with H = C. Let

X =⋃gCg−1

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so X = G. Then C nilpotent implies C ∼= T × Cu.We claim this implies that NG(T ) = NG(C). Since NG(C) ⊆ NG(T ) is tautological,

it suffices to show that NG(T ) ⊆ NG(C). Assume g ∈ NG(T ), so GTg−1 = T . ThusgTCugi

−1 = gTg−1gCug−1 = TgCug

−1. This is nilpotent, but T is the unique centraltorus, so T commutes with gCug

−1. Thus gCug−1 ⊆ ZG(T )◦ = C.

Proof to be finished next time.

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§17 April 14, 2016

(Note: was out sick the last week.)Let G be a linear algebraic group. Then we can identify the Borel subgroups of G, say

B, with the “flag variety” G/B by the map

G/B 3 gB 7→ gBg−1 ∈ B

This is a G-equivariant map, if G acts on G/B by left multiplication and on B byconjugation.

We can also do this for parabolics.

§§17.1 Vector bundles

Let V be a representation of a Borel B. Then we have a vector bundle

π : G×B V → G/B

Note that H0(G/B, V ) gives a G-representation.As a special case, let T ⊆ B be a maximal torus. Given λ ∈ X∗(T ), we have kλ an

irreducible representation of B.Now let V be a representation of G. We can restrict it to a representation of T , giving

V =⊕

λ∈X∗(T )

Vλ.

For instance, if λ ∈ x∗(T ), then we want to find which µ ∈ X∗(T ) are weights in H0(λ).

§§17.2 Interlude

For G connected reductive, we will actually see

G = BWB =⊔w∈W

BwB

for fixed B and T . Here w ∈W = W (G,T ) = NG(T )/CG(T ), and w is the lift of W andG/B =

⊔BwB/B,

There are finitely many B-orbits in bijection with W . Let U = Bu be the unipotentradical of B. In fact, as a variety U ∼= An (not as a group).

So G/B has a cellular decomposition in C, or an “etale topology” for Fq. So computingH•(G/B) or at least its dimensions is easy, determined by W and the action of W on Xand . . .

§§17.3 Weyl group

Let W = NG(T )/ZG(T ) be a Weyl group now, acting on X = X∗(T ). Convince yourselfthis makes sense, and that the action is faithful.

We can identify W with some group of automorphisms of X. Let P denote the weightsalready in g, and let P ′ denote the weights

P ′ = {α ∈ P | Gα = CG((kerα)◦) not solvable} .

If S is a subtorus of T then

W (CG(S), T ) ⊆W (G,T )

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is a subgroup, and S ⊆ Z(G). Moreover, the map G� G/S induces

W (G,T ) 'W (G/S, T/S).

Additional exercise: W is in bijection with the Borels containing T .Finally for B ⊃ T , there is a bijection

W ↔ (G/B)T

where the right-hand side are the T -fixed points in G/B, acted on by left multiplication.To see this, note that if tgB = gB ⇐⇒ gtg−1 ∈ B∀t ∈ T , then ∃b ∈ B such thatb−1g−1tgb ∈ T . So gb ∈ NG(T ) ⊃ NG(B) = B, thus g ∈ NG(T ).

Now fix α ∈ P ′, S = (kerα)◦ ⊆ Gα = ZG((kerα)◦). Then S ⊆ Z(Gα), and

Wα = W (Gα, T ) ∼= W (Gα/S, T/S).

We have T/S ∼= Gm (why is this)? Thus |Wα| ≤ 2, based on automorphisms of X∗(Gm).

Proposition 17.1

Assume G is not solvable and dimT = 1. Then |W | = 2, and dimG/B = 1.

Simple special case: look at SL2, and look at the Borel groups

B+ ={[a ∗ 0 a−1

]}B− =

{[a 0 ∗ a−1

]}.

They both contain the torus T of diagonal matrices.

Proof. Fix an isomorphism λ : Gm → T , and B ⊃ T . Recall we had a closed mapG/B → PV , and identified G/B with its image in this map. (Here V is as in the proofthat G/H is a variety.) Let φ : G→ GL(V ) be the corresponding representation.

Let e1, . . . , en be a basis of V . . . .

Example: let G = SL2. Then G/B ∼= P1, with south pole [0, 1] and north pole [1, 0],say (think of P1 as Riemann sphere). Let SL2 act on G/B in the usual fashion:[

aa−1

] [10

]=

[a0

]and [

aa−1

] [01

]=

[0a−1

].

There are two Gm-fixed points.

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§18 April 21, 2016

§§18.1 Setup

Let G be a connected linear algebraic group, and T ⊆ G a maximal torus. Let P be the

weights of T acting on g via the adjoint action, thus P ⊆ X def= X∗(T ). We let P ′ be the

subset of x ∈ P for which Gxdef= ZG ((kerx)◦) is not solvable. (We’ll actually use Gα

later, α ∈ P ′.) Eventually, we’ll start calling P ′ the “roots”.Last time, we saw that if G is not solvable and dimT = 1, then they Weyl group W

has order |W | = 2 and the dimension of the flag variety G/B is 1. The Weyl group Wacts on X∗(T ) in the same way as GLn(Z). In the situation as above, W ∼= Z/2 acts onX∗(T ) = X∗(Gm) = Z by negation; on the level of tori this means ntn−1 = t−1.

For α ∈ P ′, let Wα = W (Gα, T ) ⊆ W (G,T ) have order two and let nα ∈ NGα(T ) −ZGα(T ). Let sα be the image of nα ∈ W = W (G,T ), and let X∨ = Hom(X,Z) =X∗(T ) = Hom(Gm, T ). We get a pairing

〈, 〉X ×X∨ → Z.

We can view X ⊆ X ⊗Z R = V , and X∨ ⊆ X∨ ⊗Z R = V ∨. Thus we get a form on V ,V ∨.

Take some symmetric bilinear positive definite form, say f : V × V → R, and averageit with respect to W to get a W -invariant form

(x, y)def=∑w∈W

f(w · x,w · y)

which induces a metric on V and is invariant with respect to W .

§§18.2 Euclidean reflections

This implies sα is a Euclidean reflection, which implies

sα(x) = x− 2(α, α)−1(x, α)α.

Example 18.1

Let G = SL2(C). Then W has order two, so we can put W = (id, s). The vectorspace spanned by s has dimension 1. Get sα(α) = −α.

Here T is the diagonal matrices T = Gm =[a 0| 0 a−1

], and X = X∗(T ) = Z.

So V = X ⊗Z R = R.

Example 18.2 (A2)

Let G = SL3(C), so that T is the torus of diagonal matrices (of dimension 2), andX = X∗(T ) = Z⊕2. So V = X ⊗Z R = R2.

Now consider two-space, and let α be on the real axis, and let β be a cube rootof unity. (Thus α + β is a sixth root of unity.) So we get six roots α, −α, β, −β,α+ β, −(α+ β).

Allow reflections orthogonal to these elements. This gives S3.

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Lemma 18.3

There exists a unique α∨ ∈ V ∨ with 〈α, α∨〉 = 2 and for any x ∈ X, we have

sα(x) = x−⟨x, α∨

⟩α.

Also, if β ∈ P ′ and Gα = Gβ then sα = sβ.

Proof. First part: α∨ has ta satisfy 〈x, α∨〉 = 2(α, α)−1(x, α) for x ∈ V , which is uniquelydetermined. Second part: if Gα = Gβ we can choose nα = nβ, thus sα = sβ.

Theorem 18.4

W is generated by sα for α ∈ P ′.

Proof. Induction on dimG. Blargh.

Definition 18.5. The rank of G is defined as dimT , and its semisimiple rank is therank of G/R(G).

Lemma 18.6

We have

• G is the disjoint of B and UnB,

• R(G) = (U ∩ nUn−1)◦,

• dimU/U ∩ nUn−1 = 1.

Lemma 18.7

Assume G has semisimple rank one. Then

• dimU = 1, ZG(T ) = T , and U ∩ nUn−1 = {id}.

• There is a unique weight α of T in g such that g = gα ⊕ t ⊕ g−α, wheregα = Lie(U) and gα = Lie(nUn−1).

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Evan Chen (Spring 2016) 19 May 3, 2016

§19 May 3, 2016

Homework 6 due officially May 13. See webpage for details.

§§19.1 Some functors

Let G be a linear algebraic group, and H ⊂ G a closed subgroup. In what follows thecategory

• ModG is G-modules (allows infinite dimensional modules). (One correct way to dothis is to use a group scheme.)

• RepG is rational representations, all finite dimensional.

One can show directly that any G-module is locally finite: for all m ∈M , then kG ·m isfinite dimensional over k. (Follows by multiplicative Jordan decomposition.)

Then we have the functor

(−)H : ModG→ Vect k

which is left-exact.

Lemma 19.1

Let H, H ′ be closed subgroups in G with H ′ normalizing H. Then for M ∈ ModG,we can regard MH ∈ ModH ′.

Proof. If m ∈ MH , h ∈ H ′, we wish to check h · m ∈ MH . If h ∈ H, we haveh(h ·m) = h(h′m) = hm.

Another functor is the restriction functor

ResGH : ModG→ ModH

This is exact, and is the identity if we take the forget functor to Vect k.More interesting is the induction functor

IndGH : ModG→ ModH

It’s defined as follows. Let M ∈ ModH; then M ⊗K k[G] is a G×H module under:

• G acts trivially on M .

• G acts via the left regular representation on k[G]

• H acts as usual on M

• H acts via right regular representation.

In symbols,(g, h) · (m⊗ f) = (h ·m)⊗ ρr(h)ρ`(g)f.

Here, ρr(h)ρ`(g)f(q) = f(g−1qh).Now: 1×H is a subgroup of G×H normalized by G× 1, so we can apply the previous

lemma. In short:IndGHM

def= (M ⊗ kk[G])H ∈ ModG.

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Example 19.2

We have IndG1 k = k[G], the latter being infinite dimensional.

This functor is only left exact in general, but if H is diagonalizable then IndGH is exact.All these functors are additive and commute with direct sum, intersection, direct limits,

etc.Now, we seek the relation between Ind and Res. Recall we have the injection 1 ↪→ G,

which gives the counitεG : k[G]→ k

which is “evaluation at 1”.

Proposition 19.3 (Frobenius reciprocity)

Let M ∈ ModH, define εM : M ⊗ k[G]→M by εM = idMεG. Then

• εM induces a map IndGHM →M which is an H-module homomorphism.

• For all N ∈ ModG, then ϕ 7→ εM ◦ ϕ is an isomorphism

HomG(N, IndGHM)→ HomH(ResGH N,M).

Thus IndGH and ResGH are adjoint.

Thus, we have an isomorphism of functors

IndGH2IndH2

H1' IndGH1

from the corresponding (obvious) property for Res.

§§19.2 Root systems

From now on: G connected, reductive and T ⊂ G.Choose a system of positive roots R+ such that B+ → R+(B+); thus B is the opposite

Borel.Then W takes a system of positive roots to a system of positive roots.There exists a unique longest w0 ∈ W so that w0(R+) = −R+. Thus B is the Borel

corresponding to −r+, and B ∩B+ = t. Let Bu = U , B+u = U+.

Definition 19.4. Let λ, µ ∈ X∗(T ). Say λ ≤ µ if µ− λ is a sum of positive roots.

Let B be the Borel Since U , U+ are unipotent for any nonzero M ∈ ModG, we haveMu 6= 0 6= MU+

. Now: T normalizes U and U+ and T acts on

MU+=

⊕λ∈X∗(T )

MU+

λ .

So there exists λ, λ′ ∈ X∗(T ) with

HomB(kλ,M) 6= 0 6= HomB+(M,kλ′).

We have B � Tλ−→ Gm.

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Now if dimM <∞, then we can apply the above to M∗ to get there exists λ, λ′ suchthat

HomB(M,kλ) 6= 0 6= HomB+(M,kλ′).

Applying Frobenius reciprocity, we get

HomG(M, IndBG kλ) 6= 0 6= HomG(M, IndGB+ kλ′).

Notation: if N is a B-module, we set

H i(N) = Ri IndGB(N)

H0(N) = R0 IndGB(N) = IndGB(N).

This notation comes from sections of line bundles on G/B. Note G/T and G/B arehomotopy equivalent.

70


§20 May 12, 2016

Reference: chapters 1-8 of Humphrey’s book on category of O-modules.Let g be a semisimple over C, decomposing as

g = n− ⊕ h⊕ n.

Recall the PBW theorem tells us that ordered monomials in a basis of g correspond to abasis of

U(g) ∼= U(n−)⊕ Uh⊕ Un.

Define for λ ∈ h∗, where Cλ acts on Ub, the induced representation

IndgbCλ = Ug⊗Ub Cλ = M(λ).

We call M(λ) a Verma module, which is infinite dimensional. Thus M(λ) is a U(n−)module.

Can also deduce: there’s a mximal vector v= with U(n) · v+ = 0, and M(λ) =U(g) · v+ = U(n−) · v+. The upshot of this is that characters of M(λ) are not polynomialin eµ, µ ∈ h∗; rather we use Laurent series, which are easy to describe.

Each irreducible g-rep occurs as a quotient of Verma modules, and is the uniqueirreducible quotient; L(λ) is irreducible for each λ ∈ h∗.

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18.757 (representation of lie algebras) lecture notes · 18.757 (representation of lie algebras)...

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